The following shape is based only on squares, semicircles, and quarter circles. Find the area of the shaded part.

The Following Shape Is Based Only On Squares, Semicircles, And Quarter Circles. Find The Area Of The

Answers

Answer 1

Answer:

this? hope it helps ........

The Following Shape Is Based Only On Squares, Semicircles, And Quarter Circles. Find The Area Of The
Answer 2

Answer:

The answer is area=32pi-64 and the perimeter is 8pi

Step-by-step explanation:


Related Questions

What is the probability that a student who has no chores has a curfew ?

Answers

Answer:

15/22

Step-by-step explanation:

Of the 66 students who have no chores, 45 have a curfew.  So the probability is 45/66 = 15/22.

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

21, 14, 13, 24, 17, 22, 25, 12

Required:
a. Calculate the sample mean and the sample standard deviation.
b. Construct the 90% confidence interval for the population mean.
c. Construct the 95% confidence interval for the population mean

Answers

Answer:

a

   [tex]\= x = 18.5[/tex]  ,  [tex]\sigma = 5.15[/tex]

b

 [tex]15.505 < \mu < 21.495[/tex]

c

 [tex]14.93 < \mu < 22.069[/tex]

Step-by-step explanation:

From the question we are are told that

    The  sample data is  21, 14, 13, 24, 17, 22, 25, 12

     The sample size is  n  = 8

Generally the ample mean is evaluated as

        [tex]\= x = \frac{\sum x }{n}[/tex]

        [tex]\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}[/tex]

         [tex]\= x = 18.5[/tex]

Generally the standard deviation is mathematically evaluated as

         [tex]\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}[/tex]

[tex]\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}[/tex]

[tex]\sigma = 5.15[/tex]

considering part b

Given that the confidence level is  90% then the significance level is evaluated as

         [tex]\alpha = 100-90[/tex]

         [tex]\alpha = 10\%[/tex]

         [tex]\alpha = 0.10[/tex]

Next we obtain the critical value of  [tex]\frac{ \alpha }{2}[/tex]  from the normal distribution table the value is  

     [tex]Z_{\frac{ \alpha }{2} } = 1.645[/tex]

The margin of error is mathematically represented as

      [tex]E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]

=>    [tex]E =1.645 * \frac{5.15 }{\sqrt{8} }[/tex]

=>     [tex]E = 2.995[/tex]

The 90% confidence interval is evaluated as

       [tex]\= x - E < \mu < \= x + E[/tex]

substituting values

       [tex]18.5 - 2.995 < \mu < 18.5 + 2.995[/tex]

       [tex]15.505 < \mu < 21.495[/tex]

considering part c

Given that the confidence level is  95% then the significance level is evaluated as

         [tex]\alpha = 100-95[/tex]

         [tex]\alpha = 5\%[/tex]

         [tex]\alpha = 0.05[/tex]

Next we obtain the critical value of  [tex]\frac{ \alpha }{2}[/tex]  from the normal distribution table the value is  

     [tex]Z_{\frac{ \alpha }{2} } = 1.96[/tex]

The margin of error is mathematically represented as

      [tex]E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]

=>    [tex]E =1.96 * \frac{5.15 }{\sqrt{8} }[/tex]

=>     [tex]E = 3.569[/tex]

The 95% confidence interval is evaluated as

       [tex]\= x - E < \mu < \= x + E[/tex]

substituting values

       [tex]18.5 - 3.569 < \mu < 18.5 + 3.569[/tex]

       [tex]14.93 < \mu < 22.069[/tex]

Answer two questions about Equations A and B: A.5x=20 \ B.x=4 ​ 1) How can we get Equation B from Equation A? Choose 1 answer: (Choice A) Multiply/divide both sides by the same non-zero constant (Choice B,) Multiply/divide both sides by the same variable expression (Choice C) Add/subtract the same quantity to/from both sides (Choice D) Add/subtract a quantity to/from only one side

Answers

Answer:

Multiply/divide both sides by the same non-zero constant

Step-by-step explanation:

5x = 20

Divide each side by 5

5x/5 = 20/5

x = 4

To obtain (B) from (A) "Multiply/divide both sides by the same non-zero constant"

Given the equations :

5x = 20 ___ (A)x = 4 _____ (B)

To obtain the value ; x = 4 from A

We multiply (A) by the same non-zero constant

Here, the constant value which can be used is 5 in other to isolate 'x'

5x/5 = 20/5

x = 4

Therefore, to obtain (B) from (A) "Multiply/divide both sides by the same non-zero constant"

Learn more on equations :https://brainly.com/question/2972832

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4 solid cubes were made out of the same material. All four have different side lengths: 6cm, 8cm, 10cm, and 12cm. How to distribute the cubes onto two plates of a scale so the scale is balanced? Answer: A= the cube with side length 6 cm, B= the cube with side length 8 cm, C= the cube with side length 10 cm, D= the cube with side length 12 cm. On one side of the scale : , on the other side of the scale

Answers

Answer: The cube with side length of 12cm is alone in one plate, the other 3 cubes are in the other plate.

Step-by-step explanation:

We have 4 cubes with side lengths of:

6cm, 8cm, 10cm and 12cm.

Now, some things you need to know:

If we want a scale to be balanced, then the mass in both plates must be the same.

The volume of a cube of side length L is:

V = L^3

And the mass of an object of density D, and volume V is:

M = D*V.

As all the cubes are of the same material, all of them have the same density, so the fact that we do not know the value of D actually does not matter here.

Then we want to forms two groups of cubes in such a way that the total volume in each plate is the same (or about the same), the volumes of the cubes are:

Cube of 6cm:

V = (6cm)^3 = 216cm^3

Cube of 8cm:

V = (8cm)^3 = 512cm^3

Cube of 10cm:

V = (10cm)^3 = 1000cm^3

cube of 12cm

V = (12cm)^3 =  1728cm^3

First, if we add the volumes of the first two cubes, we have:

V1 = 216cm^3 + 512cm^3 = 728cm^3

Now we can see that we add 1000cm^3 the volume will be equal to the volume of the larger cube, so here we can also add the cube with side length of 10cm

Then the volume of the 3 smaller cubes together is:

V1 = 216cm^3 + 512cm^3 + 1000cm^3 = 1728cm^3.

Then, if we want to have the same volume in each plate, then we need to have the 3 smaller cubes in one plate, and the larger cube in the other plate.

What is the solution of the system of equations?
y = -3x + 7
y = 2x - 8

Answers

Answer:

x = 3, y = -2

Step-by-step explanation:

Since y=y

then, -3x +7 = 2x-8

7+8 = 3x+2x

15 = 5x

x=3

substitute

y = 2(3) - 8

y = -2

Hope that helped!!! k

Answer:

y = -2

x = 3

Step-by-step explanation:

Solve using elimination

1. Rearrange the equations to make it easier to solve

y = -3x + 7 → 3x + y = 7

y = 2x - 8 → 2x - y = 8

2. Multiply the equations to have a matching coefficient

2(3x + y = 7) = 6x + 2y = 14

3(2x - y = 8) = 6x - 3y = 24

3. Subtract

 6x + 2y = 14

- 6x - 3y = 24

  0 + 5y = -10

4. Solve for y

5y = -10

y = -2

5. Substitute y in any equation to solve for x

-2 = -3x + 7

-3x = -9

x = 3

An arithmetic sequence has this recursive formula: (a^1 =8, a^n= a^n-1 -6
A.a^n=8+(n-6)(-1)
B.a^n=8+(n-1)(-6)
C.

Answers

Answer:

[tex]a_n = 8 + (n - 1) (-6)[/tex]

Step-by-step explanation:

Given

[tex]a_1 = 8[/tex]

Recursive: [tex]a_{n} = a_{n-1} - 6[/tex]

Required

Determine the formula

Substitute 2 for n to determine [tex]a_2[/tex]

[tex]a_{2} = a_{2-1} - 6[/tex]

[tex]a_{2} = a_{1} - 6[/tex]

Substitute [tex]a_1 = 8[/tex]

[tex]a_2 = 8 - 6[/tex]

[tex]a_2 = 2[/tex]

Next is to determine the common difference, d;

[tex]d = a_2 - a_1[/tex]

[tex]d = 2 - 8[/tex]

[tex]d = -6[/tex]

The nth term of an arithmetic sequence is calculated as

[tex]a_n = a_1 + (n - 1)d[/tex]

Substitute [tex]a_1 = 8[/tex] and [tex]d = -6[/tex]

[tex]a_n = a_1 + (n - 1)d[/tex]

[tex]a_n = 8 + (n - 1) (-6)[/tex]

Hence, the nth term of the sequence can be calculated using[tex]a_n = 8 + (n - 1) (-6)[/tex]

pls help :Find the missing side or angle.
Round to the nearest tenth.

Answers

Answer:

C° = 71.6056

Step-by-step explanation:

Law of Cosines: c² = a² + b² - 2abcosC°

Step 1: Plug in known variables

29² = 30² + 15² - 2(30)(15)cosC°

Step 2: Evaluate

841 = 900 + 225 - 900cosC°

-59 = 225 - 900cosC°

-284 = -900cosC°

71/225 = cosC°

cos⁻¹(71/225) = C°

C° = 71.6056

And we have our answer!

Answer:

  79.0°

Step-by-step explanation:

The Law of Cosines is used for this purpose. It tells you ...

  a² = b² +c² -2bc·cos(A)

Solving for A gives ...

  cos(A) = (b² +c² -a²)/(2bc) = (15² +29²-30²)/(2(15)(29)) = 166/870

Using the inverse cosine function, we find the angle to be ...

  A = arccos(166/870) ≈ 79.00026°

  A ≈ 79.0°

what number should replace the question mark

Answers

Answer: The missing number is 5.

Step-by-step explanation:

In the table we can only have numbers between 1 and 9,

The pattern that i see is:

We have sets of 3 numbers.

"the bottom number is equal to the difference between the two first numers, if the difference is negative, change the sign, if the difference is zero, there goes a 9 (the next number to zero)"

Goin from right to left we have:

9 - 6 = 3

6 - 2 = 4

4 - 9 = - 5 (is negative, so we actually use -(-5) = 5)

4 - 4 = 0 (we can not use zero, so we use the next number, 9)

3 - 3 = 0 (same as above)

? - 1 = 4

? = 4 + 1 =  5

The missing number is 5.

Techwiz electronics makes a profit of $35 for each mp3 and $18 for each DVD last week techwiz sold a combined total of 118 mp3 and DVD players. Let x be the number of mp3 sold last week write an expression for the combined total profit (in dollars) made last week

Answers

Answer:

The total   profit is [tex]p = 17x + 2124[/tex]

Step-by-step explanation:

From the question we are told that

   The profit made on each mp3 is  k  =  $35

    The profit made on each mp3 is  y =  $18

     The total amount sold is   n  =  118

 Now given that the amount of mp3 sold is x then the amount of  DVD sold is mathematically evaluated as

                      [tex]n - x[/tex]

Now the profit made on the x number of mp3 sold is  

                      [tex]x * 35 = 3x[/tex]

And the the profit made from the n-x number of  DVD  sold is  18 (n-x ) =  18 - 18x

 So the total profit made last week from the sales of both mp3 and DVD  is  

                   [tex]p = 35x + 18n - 18x[/tex]

                    [tex]p = 17x + 18(118)[/tex]

                    [tex]p = 17x + 2124[/tex]

Find the sum. 31.25 + 9.38

Answers

Answer:

40.63

Step-by-step explanation:

31.25+9.38= 40.63

Hope this helps

Answer: 40.63

Look at the image for shown work.

if the LCM and the HCF of two numbers are 9 and 3, respectively, what are the numbers?

Answers

Hey  There!

Answer:

HCF = 9  (With the two numbers) - 18,9LCM = 3  (with the two numbers) -  6,9

Step-by-step explanation:

HCF

If HCF is  ''9'' that means that ''9'' is the divisible of two numbers.

So 18 and 19 can be divided by 9 and that's the highest divisible for both factors.

And always remeber the answer is a ''Prime factor.''

LCM

If LCM is ''3'' that means ''3'' is the lowest common multiple out of  two numbers.

Hope this helps!

Have a nice Day!:)

How would you find the coefficient of the third term in (x+5)^7?

Answers

Answer:

The answer is option B

Step-by-step explanation:

To find the coefficient of the third term in

[tex](x + 5)^{7} [/tex]

Rewrite the expansion in the form

[tex](a + x)^{n} [/tex]

where n is the index

So we have

[tex] ({5 + x})^{7} [/tex]

After that we use the formula

[tex]nCr( {a}^{n - r} ) {x}^{r} [/tex]

where r is the term we are looking for

For the third term we are looking for the term containing x²

that's

r + 1 = 3

r = 2

So to find the coefficient of the third term

We have

[tex]7C2[/tex]

Hope this helps you

first of all, the notation is wrong it should be [tex] {}^nC_r \text{ and more usual notation is } {n \choose k} [/tex]

second, the

[tex](r+1)^{\text{th}} \text{ term } T_{r+1} \text{ in the expansion of } (x+a)^n \text{ is } {n \choose r}x^{(n-r)}a^r[/tex]

here [tex] a=5 \text{ and } n=7 \text{ and for } 3^{\text{rd}} \text{ term } T_3, \quad r+1=3 \implies r=2 [/tex]

so the coefficient of third term is, [tex]{7 \choose 2}={7\choose 5}[/tex]

an important property of binomial coefficient you should know:

[tex] {n \choose k}={n \choose {n-k}}[/tex]

and if you interchange [tex] x \text{ and } a[/tex]

only the "order" will get reversed. i.e. the series will start from back.

another thing, the [tex] k^{\text{th}} \text{ term from beginning, is the } (n-k+2)^{\text{th}} \text{ term from behind}[/tex]

Please helppp meee I don’t know the answer

Answers

Answer:

First use the protractor then round the number to the nearest 10

Answer:

Round to the nearest tenth

Step-by-step explanation:

In parallelogram PQSR, what is PQ? 2 cm 5 cm 6 cm 9 cm

Answers

Answer:

D) 9 cm

Step-by-step explanation:

EDGE 2020

(D) 9 cm.

Parallelogram:A simple (non-self-intersecting) quadrilateral with two sets of parallel sides is known as a parallelogram in Euclidean geometry. A parallelogram's facing or opposing sides are of equal length, and its opposing angles are of similar size. The Euclidean parallel postulate or one of its equivalent formulations must be used in order to demonstrate the congruence of opposed sides and opposite angles because both conditions are a direct result of this postulate.In contrast, a quadrilateral with only one set of parallel sides is referred to as a trapezoid or trapezium in British or American English.The parallelepiped is a parallelogram's three-dimensional equivalent.

Therefore, the correct answer is (D) 9 cm.

Know more about a parallelogram here:

https://brainly.com/question/970600

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Multiple-Choice Questions
1. In 1995, Diana read 10 English books and 7 French books. In 1996, she read twice as many French books as English books. If 60% of the books that she read during the 2 years were French, how many English and French books did she read in 1996?
(A) 16
(B) 26
(0) 32
(D) 48​

Answers

Answer:

(D) 48​

Step-by-step explanation:

Let English book = x

Let french book = y

In 1995 x= 10

Y= 7

In 1996

Y = 2x

Total book read in the two years

0.6(Total) = y

0.4(total) = x

We don't know the exact amount of books read in 1996.

Total = 10 + 7 +x +2x

Total = 17+3x

0.6(total) = 7+2x

0.6(17+3x) = 7+2x

10.2 +1.8x= 7+2x

10.2-7= 2x-1.8x

3.2= 0.2x

3.2/0.2= x

16= x

So she read 16 English book

And 16*2 = 32 french book Making it a total of 16+32= 48 books in 1996

Please help!! find the circumference of a circle with a diameter of 13 meters

Answers

Answer:

C = 2pie(r)

r= d/2= 13/2= 6.5

C = 2*3.14*6.5

C= 41

Step-by-step explanation:

ΔABC is similar to ΔMNO. The scale factor from ΔMNO to ΔABC is 3∕2 . If the area of ΔMNO is 10 square units, what's the area of ΔABC? Question 12 options: A) 45 square units B) 90 square units C) 22.5 square units D) 15 square units

Answers

Answer:

The area of ΔABC= 6.667 square units

Step-by-step explanation:

ΔABC is similar to ΔMNO.

The scale factor from ΔMNO to ΔABC is 3∕2

the area of ΔMNO is 10 square units,

The area of ΔABC/the area of ΔMNO

= 2/3

The area of ΔABC/10= 2/3

The area of ΔABC= 2/3 * 10

The area of ΔABC= 20/3

The area of ΔABC= 6 2/3

The area of ΔABC= 6.667 square units

Answer:

22.5 square units

Step-by-step explanation:

i multiplied 10 by 2 to get 20 and went with the closest answer and got it right.

i dont know how to do math but i guess it worked

2/5×1 3/12? plz help meh​

Answers

Answer:

[tex]\boxed{\frac{1}{2} }[/tex]

Step-by-step explanation:

Hey there!

Well given,

[tex]\frac{2}{5} * 1 \frac{3}{12}[/tex]

We need to make 1 3/12 improper,

1*12 = 12

12 + 3 = 15

[tex]\frac{2}{5} * \frac{15}{12}[/tex]

2*15 = 30

5*12 = 60

[tex]\frac{30}{60}[/tex]

Simplified

[tex]\frac{1}{2}[/tex]

Hope this helps :)

The joint density function for a pair of random variables X and Y is given. f(x, y) = Cx(1 + y) if 0 ≤ x ≤ 4, 0 ≤ y ≤ 4 0 otherwise f(x,y) = 0
A) Find the value of the constant C. I already have 1/24.
B) Find P(X < = 1, Y < = 1)
C) Find P(X + Y < = 1).

Answers

Answer:

A) C = 1/96

B) P(x<=1, y<=1) = 1/128 or 0.0078125 to 7 places

C) P(x+y<=1) = 5/2305, or 0.0021701 to 7 places

Step-by-step explanation:

f(x,y) = C x (1+y)

A)

To find C, we need to integrate the volume under region bound by

0 <= x <= 4, and

0 <= y <= 4

This volume equals 1.0.

Find integral,

int( int(f(x,y),x=0,4), y = 0,4) = 96C

therefore C = 1/96

or

F(x,y) = x (1+y) / 96  ............................(1)

B)

P(x<=1, y<=1)

Repeat the integral, substitute the appropriate limits,

P = int( int(F(x,y),x=0,1), y = 0,1)

= 1/128 or 0.0078125

P(x<=1, y<=1) = 1/128 or 0.0078125 to 7 places

C)

P(x+y<=1)

From the function, we know that this is going to be less than one half of the probability in (B), closer to 1/4 of the previous.

It will be again a double integral, as follows:

P = int( int(F(x,y),x=0,1-y), y = 0,1)

= 5/2304

= 0.0021701 (to 7 decimals)

P(x+y<=1) = 5/2305, or 0.0021701 to 7 places

a student ran out of time on a multiple choice exam and randomly guess the answers for two problems each problem have four answer choices ABCD and only one correct answer what is the probability that he answered neither of the problems correctly ​

Answers

Answer:

The probability that he answered neither of the problems correctly ​is 0.0625.

Step-by-step explanation:

We are given that a student ran out of time on a multiple-choice exam and randomly guess the answers for two problems each problem have four answer choices ABCD and only one correct answer.

Let X = Number of problems correctly ​answered by a student.

The above situation can be represented through binomial distribution;

[tex]P(X=r)=\binom{n}{r}\times p^{r}\times (1-p)^{n-r};x=0,1,2,3,....[/tex]    

where, n = number of trials (samples) taken = 2 problems

           r = number of success = neither of the problems are correct

           p = probability of success which in our question is probability that

                 a student answer correctly, i.e; p = [tex]\frac{1}{4}[/tex] = 0.75.

So, X ~ Binom(n = 2, p = 0.75)

Now, the probability that he answered neither of the problems correctly ​is given by = P(X = 0)

             P(X = 0) = [tex]\binom{2}{0}\times 0.75^{0}\times (1-0.75)^{2-0}[/tex]

                            = [tex]1 \times 1\times 0.25^{2}[/tex]

                            = 0.0625

Please help. I’ll mark you as brainliest if correct!

Answers

Answer:

x and y can have many values

Step-by-step explanation:

-24x - 12y = -16

Then: 24x + 12y = 16

We know: 6x + 3y = 4

X and Y can have a lot of valoues.

6x + 3y = 4

3 ( 2x + y) = 4

2x + y= 4/3

2x+y= 1.333...

Find the side length, b.
Round to the nearest tenth.

Answers

Answer:

b ≈ 9.2

Step-by-step explanation:

Using Pythagoras' identity in the right triangle.

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

b² = a² + c² = 6² +7² = 36 + 49 = 85 ( take the square root of both sides )

b = [tex]\sqrt{85}[/tex] ≈ 9.2 ( to the nearest tenth )

Answer:

9.22

Step-by-step explanation:

Since it's a 90° triangle [tex]c^{2} =a^{2} +b^{2}[/tex].

In this example they labeled the hypotenuse as b instead of c are equation is still the same just put the correct variables in the right places.

[tex]b = \sqrt{6^{2} +7^{2} }[/tex]

b = 9.22

On a particular production line, the likelihood that a light bulb is defective is 10%. seven light bulbs are randomly selected. What is the probability that at most 4 of the light bulbs will be defective

Answers

Answer:

0.9995

Step-by-step explanation:

10% = 0.10

1 - 0.10 = 0.9

n = number of light bulbs = 7

we calculate this using binomial distribution.

p(x) = nCx × p^x(1-p)^n-x

our question says at most 4 is defective

= (7C0 × 0.1⁰ × 0.9⁷) + (7C1 × 0.1¹ × 0.9⁶) + (7C2 × 0.1² × 0.9⁵) + (7C3 × 0.1³ × 0.9⁴) + (7C4 × 0.1⁴ × 0.9³)

= 0.478 + 0.372 + 0.1239 + 0.023 + 0.0026

= 0.9995

we have 0.9995 probability that at most 4 light bulbs are defective.

A. Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 1; B: the numbers add to 4; C: at least one of the numbers is 1; and D: the numbers do not add to 10. Express the given event in symbols.
The red die shows 1 and the numbers add to 4.
How many elements does it contain?
B. Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 3; B: the numbers add to 2; C: at least one of the numbers is 1; and D: the numbers do not add to 10. Express the given event in symbols. HINT [See Example 5.]
The numbers do not add to 2.
How many elements does it contain?
C. Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 1; B: the numbers add to 2; C: at least one of the numbers is 3; and D: the numbers do not add to 11. Express the given event in symbols. HINT [See Example 5.]
Either the numbers add to 11 or the red die shows a 1.
How many elements does it contain?
D. Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 4; B: the numbers add to 5; C: at least one of the numbers is 1; and D: the numbers do not add to 9. Express the given event in symbols. HINT [See Example 5.]
Either the numbers add to 5, or they add to 9, or at least one of them is 1.
How many elements does it contain?

Answers

Answer:

1. elements it contains =  (1,3)

2.  elements it contains = 35

3.  elements it contains = 8

4.  elements it contains = 17

Step-by-step explanation:

A. Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 1; B: the numbers add to 4; C: at least one of the numbers is 1; and D: the numbers do not add to 10. Express the given event in symbols.

The red die shows 1 and the numbers add to 4.

How many elements does it contain?

B. Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 3; B: the numbers add to 2; C: at least one of the numbers is 1; and D: the numbers do not add to 10. Express the given event in symbols. HINT [See Example 5.]

The numbers do not add to 2.

How many elements does it contain?

C. Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 1; B: the numbers add to 2; C: at least one of the numbers is 3; and D: the numbers do not add to 11. Express the given event in symbols. HINT [See Example 5.]

Either the numbers add to 11 or the red die shows a 1.

How many elements does it contain?

D. Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 4; B: the numbers add to 5; C: at least one of the numbers is 1; and D: the numbers do not add to 9. Express the given event in symbols. HINT [See Example 5.]

Either the numbers add to 5, or they add to 9, or at least one of them is 1.

How many elements does it contain?

NB. Attached is the solution to the problems stated above

I need help please help meee I don’t understand

Answers

Answer:

204

Step-by-step explanation:

To simplify the shape, you can do multiple things. I've opted to shave down both prongs to take it from a 'T' shape to a rectangular prism.

For height of the prongs, take 4 from 6.

6 - 4 = 2

Divide by 2 as there are 2 prongs.

2 / 2 = 1

Remember L * W * H

6 * 3 * 1 = 18

Remember that there are two prongs!

3 + 4 = 7

6 * 7 * 4 = 168

168 + 2(18) = 204

What are the solution(s) of the quadratic equation 98 - x2 = 0?
x = +27
Ox= +63
x = +7/2
no real solution

Answers

Answer:

±7 sqrt(2) = x

Step-by-step explanation:

98 - x^2 = 0

Add x^2 to each side

98 =x^2

Take the square root of each side

±sqrt(98) = sqrt(x^2)

±sqrt(49*2) = x

±7 sqrt(2) = x

Answer:

[tex]\huge \boxed{{x = \pm 7\sqrt{2} }}[/tex]

Step-by-step explanation:

[tex]98-x^2 =0[/tex]

[tex]\sf Add \ x^2 \ to \ both \ sides.[/tex]

[tex]98=x^2[/tex]

[tex]\sf Take \ the \ square \ root \ of \ both \ sides.[/tex]

[tex]\pm \sqrt{98} =x[/tex]

[tex]\sf Simplify \ radical.[/tex]

[tex]\pm \sqrt{49} \sqrt{2} =x[/tex]

[tex]\pm 7\sqrt{2} =x[/tex]

[tex]\sf Switch \ sides.[/tex]

[tex]x= \pm 7\sqrt{2}[/tex]

If tanA = 3
evaluate
CosA + sinA\
casA - SinA​

Answers

Answer:

Hi, there!!!

I hope you mean to evaluate cosA+ sonA /cosA - sinA.

so, i hope the answer in pictures will help you.

In a recent​ year, the scores for the reading portion of a test were normally​ distributed, with a mean of and a standard deviation of . Complete parts​ (a) through​ (d) below. ​(a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than . The probability of a student scoring less than is nothing. ​(Round to four decimal places as​ needed.) ​(b) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is between and . The probability of a student scoring between and is nothing. ​(Round to four decimal places as​ needed.) ​(c) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is more than . The probability of a student scoring more than is nothing. ​(Round to four decimal places as​ needed.) ​(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below. A. than 0.05. B. than 0.05. C. The event in part is unusual because its probability is less than 0.05. D. The events in parts are unusual because its probabilities are less than 0.05.

Answers

The question is incomplete. Here is the complete question.

In a recent year, the socres for the reading portion of a test were normally distributed, with a mean of 23.3 and a standard deviation of 6.4. Complete parts (a) through (d) below.

(a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 18. (Round to 4 decimal places as needed.)

(b) Find a probability that a random selected high school student who took the reading portion of the test has a score that is between 19.9 and 26.7.

(c) Find a probability that a random selected high school student who took the reading portion of the test ahs a score that is more than 36.4.

(d) Identify any unusual events. Explain your reasoning.

Answer: (a) P(X<18) = 0.2033

              (b) P(19.9<X<26.7) = 0.4505

              (c) P(X>36.4) = 0.0202

               (d) Unusual event: P(X>36.4)

Step-by-step explanation: First, determine the z-score by calculating:

[tex]z = \frac{x-\mu}{\sigma}[/tex]

Then, use z-score table to determine the values.

(a) x = 18

[tex]z = \frac{18-23.3}{6.4}[/tex]

z = -0.83

P(X<18) = P(z< -0.83)

P(X<18) = 0.2033

(b) x=19.9 and x=26.7

[tex]z = \frac{19.9-23.3}{6.4}[/tex]

z = -0.67

[tex]z = \frac{26.7-23.3}{6.4}[/tex]

z = 0.53

P(19.9<X<26.7) = P(z<0.53) - P(z< -0.67)

P(19.9<X<26.7) = 0.7019 - 0.2514

P(19.9<X<26.7) = 0.4505

(c) x=36.4

[tex]z = \frac{36.4-23.3}{6.4}[/tex]

z = 2.05

P(X>36.4) = P(z>2.05) = 1 - P(z<2.05)

P(X>36.4) = 1 - 0.9798

P(X>36.4) = 0.0202

(d) Events are unusual if probability is less than 5% or 0.05. So, part (c) has an unusual event.

The probability will be:

(a) 0.2038

(b) 0.4046

(c) 0.0203

(d) Event in part (c) is unusual.

According to the question,

[tex]\mu = 23.2[/tex][tex]\sigma = 6.4[/tex]

Let,

"X" shows the test scores.

(a)

The z-score for X=18 will be:

→ [tex]z = \frac{X- \mu}{\sigma}[/tex]

      [tex]= \frac{18-23.3}{6.4}[/tex]

      [tex]= -0.828[/tex]

So,

The probability will be:

→ [tex]P(X<18) = P(z < -0.828)[/tex]

                     [tex]= 0.2038[/tex]

(b)

The z-score for X=19.9 will be:

→ [tex]z = \frac{X -\mu}{\sigma}[/tex]

     [tex]= \frac{19.9-23.3}{6.4}[/tex]

     [tex]= -0.531[/tex]

The z-score for X=26.7 will be:

→ [tex]z = \frac{X -\mu}{\sigma}[/tex]

      [tex]= \frac{26.7-23.3}{6.4}[/tex]

      [tex]= 0.531[/tex]

So,

The probability will be:

→ [tex]P(19.9 < X< 23.3) = P(-0.531 < z< 0.531)[/tex]

                                   [tex]= 0.4046[/tex]

(c)

The z-score for X=36.4 will be:

→ [tex]z = \frac{X -\mu}{\sigma}[/tex]

     [tex]= \frac{36.4-23.3}{6.4}[/tex]

     [tex]= 2.047[/tex]

So,

The probability will be:

→ [tex]P(X > 36.4 )= P(z > 2.047)[/tex]

                       [tex]= 0.0203[/tex]

(d)

Just because it's probability value is less than 0.05, so that the events is "part c" is unusual.

Learn more about probability here:

https://brainly.com/question/23044118

Please help. I’ll mark you as brainliest if correct! Thank you

Answers

Answer:

8 pounds of cheaper candy,

17.5 pounds of expensive candy

Step-by-step explanation:

Let's define some variables. Let's say the amount of pounds of candy that sells for $2.20/lb is x, and the $7.30 is y. Now we can write some equations!

x + y = 25.5

[tex]\frac{2.2x + 7.3y}{25.5} = 5.7[/tex]

We can start substitution. We can say that x = 25.5 - y. Plugging this into our second equation, we get:

y = 17.5

Plugging this in, we find that:

x = 8.

Match each function name with its equation.

Answers

Answer:

a. Quadratic--[tex]y=x^{2}[/tex]

b. Absolute Value--[tex]y=|x|[/tex]

c. Linear--[tex]y=x[/tex]

d. Reciprocal Squared--[tex]y=\frac{1}{x^{2} }[/tex]

e. Cubic--[tex]y=x^{3}[/tex]

f. Square Root--[tex]y=\sqrt{x}[/tex]

g. Reciprocal--[tex]y=\frac{1}{x}[/tex]

h. Cube root--[tex]y=\sqrt[3]{x}[/tex]

Answer:

Step-by-step explanation:

y=[tex]x^{2}[/tex] is quadratic

y=x  is an absolute value

y= |x| os linear

y= [tex]\frac{1}{x}[/tex] is reciprocal

y= [tex]x^{3}[/tex] is cubic

y= [tex]\sqrt{x}[/tex] is square root

y= [tex]\frac{1}{x^{2} }[/tex] is reciprocal squared

y= [tex]\sqrt[3]{x}[/tex] is cube root

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