When x€Q, what is the solution of 3x-2/2=x-1/2 ?​

Answer:

16

Step-by-step explanation:

The sequence is 1, 4,...,64, 256, 1024

Notice that:

● 1 = 2^0

● 4 = 2^2

● 64 = 2^6

● 256 = 2^8

● 1024 = 2^10

Notice that we add 2 each time to the exponent so the missing number is:

● 2^(2+2) = 2^4 = 16

Complete Question

On the uploaded image is a similar question that will explain the given question

Answer:

The value of k is  [tex]k = 214285.7[/tex]

The percentage  of the oil that will be cleaned is [tex]x = 80.77\%[/tex]

Step-by-step explanation:

From the question we are told that

   The  cost of cleaning up the spillage is  [tex]C = \frac{ k x }{100 - x }[/tex]  [tex]x \le x \le 100[/tex]

     The  cost of cleaning x =  70% of the oil is  [tex]C = \$500,000[/tex]

Now at  [tex]C = \$500,000[/tex] we have  

       [tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]

       [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]k = 214285.7[/tex]

Now  When  [tex]C = \$900,000[/tex]

       [tex]x = 80.77\%[/tex]

There will be a total of 4 test scores including the final exam. To get a 70, the 4 tests need to equal 4 x 70 = 280 points , to be 79, they have to equal 4 x 79 = 316 points.

The 3 already done = 61 + 62 + 86 = 209 points.

The final exam needs to be between :

280 -209 = 71

316 -209 = 107. The answer would be between 71 and 100%

Answer:

Solution : Option B

Step-by-Step Explanation:

We have the following system of equations at hand here.

{ x = 5 cot(t), y = - 3csc(t) + 4 }

Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,

x = 5 cot(t) ⇒ x - 5 = cot(t),

y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)

Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations  as well. --- Step #2

( y - 4 / - 3 )² = (csc(t))²

- ( x - 5 / 1 )² = (cot(t))²  

___________________

(y - 4)² / 9 - x² / 25 = 1

And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.

Answer:

A.) x=90°

Step-by-step explanation:

Note:

The triangle shown is an isosceles triangle, which means that it has 2 congruent sides (as shown by the small intersecting lines), and this also means that it has two congruent angles.

We are given an angle measure adjacent to one of the missing angles. These two form supplementary angles, which means that they're sum is equal to 180°, or a straight line. So, to find:

[tex]180=135+y[/tex]

y is the unknown angle. Solve for y:

[tex]180-135=y\\\\y=45[/tex]

y is 45°. Since this and the other angle are congruent, add:

[tex]45+45=90[/tex]

Note:

Triangles angles will always add up to a total of 180°.

To find the missing angle x°, use:

[tex]180=a+b+c[/tex]

These are the angles in a triangle. Substitute any known values and solve:

[tex]180=45+45+x\\\\180=90+x\\\\180-90=x\\\\x=90[/tex]

The missing angle x° is 90°.

:Done

Answer:

Option (D)

Step-by-step explanation:

Formula to get the area of a regular polygon in a circle will be,

Area = [tex]n[\frac{1}{2}\times (\text{Base})\times (\text{Height})][/tex]

        = [tex]n[\frac{1}{2}\times (\text{s})\times (\text{h})][/tex]

Here 'n' is the number of sides.

If n increases, h approaches r so that 'rh' approaches r².

In other words, if the number of sides of the polygon gets increased, area of the polygon approaches the area of the circle.

Therefore, Option (4) will be the answer.

In this exercise it is necessary to have knowledge about polygons, so we have to:

Letter D

Then using the formula for the area of ​​a regular polygon we find that:

[tex]A=n(1/2*B*H)\\=n(1/2*S*H)[/tex]

So from this way we were not able to identify the option that best corresponds to this alternative.

See more about polygons at  brainly.com/question/17756657

Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]

$a^{2}=5+2 \sqrt{6}$

$a^{3}=11 \sqrt{2}+9 \sqrt{3}$

The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.

Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$

so fits with the other answers.

Answer:

[tex]y^3 -6y-6[/tex]

Answer:

1048576

Step-by-step explanation:

Given the following :

Pizza order :

Size = small, medium, large, or extra large = 4 possible sizes

Toppings = any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8).

Combination of Toppings = 2^8

Four different sizes of pizza = 4

Number of possibilities in ordering for a single pizza :

(4 * 2^8) = 4 * 256 = 1024

Number of possibilities in ordering two pizzas :

(4 * 2^8)^2

(2^2 * 2^8)^2

From indices :

[2^(2+8)]^2

[2^(10)]^2

2^(10*2)

2^20

= 1048576

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Answer:

Question 1 = D) Acute

Question 2 = C)3 feet

Question 3 = D) Obtuse

Question 4 = C)27.73 ft.

Step-by-step explanation:

Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths

5 = a, 6 = b and c = 7

a² + b² = c²

5² + 6² = 7²

25 + 36 = 49

61 = 49

61 ≠ 49, Hence 61 > 49

Therefore, this is an Acute Triangle

Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?

This is question that deals with proportion.

The formula to solve for this:

Height of the statue/ Length of the shadow of the person = Height of the person/ Length of the shadow of the person

Height of the statue = 15 feet

Length of the shadow of the person = 20 feet

Height of the person = unknown

Length of the shadow of the person = 4

15/ 20 = Height of the person/4

Cross Multiply

15 × 4 = 20 × Height of the person

Height of the person = 15 × 4/20

= 60/20

Height of the person = 3 feet

Therefore, the person is 3 feet tall.

Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths 17, 12, 9

9 = a, 12 = b and c = 17

a² + b² = c²

9² + 12² = 17²

81 + 144 = 289

225 = 289

225 ≠ 289

225 < 289

Hence, This is an Obtuse Triangle.

Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?

To calculate how far apart the two friends are we use the formula

Distance = √ ( Length² + Breadth²)

We are given dimensions: 12ft by 25ft

Length = 12ft

Breadth = 25ft

Distance = √(12ft)² + (25ft)²

Distance = √144ft²+ 625ft²

Distance = √769ft²

Distance = 27.730849248ft

Approximately ≈27.73ft

Therefore, the friends are 27.73ft apart.

Answer: y=(4/3)x+2/3

Step-by-step explanation:

Slope-intercept form is expressed as y=mx+b

First, find the slope (m):

m= rise/run or vertical/horizontal or y/x (found between the highlighted points)

m = 4/3

Second, find b:

Use one of the highlighted points for (x, y)

2=4/3(1)+b

6/3=4/3+b

2/3=b

b=2/3

Plug it into the equation:

You get y=(4/3)x+2/3 :)

Answer:

50.93

Step-by-step explanation:

Add up the frequencies:

2 + 5 + 14 + 15 + 21 + 18 + 15 + 9 + 2 = 101

Divide by 2: 101/2 = 50.5

So the median is the 51st number, with 50 below and 50 above.

Add up the frequencies until you find the interval that contains the 51st number.

2 + 5 + 14 + 15 = 36

2 + 5 + 14 + 15 + 21 = 57

So the median is in the group 49.5 − 51.5.  To estimate the median, we use interpolation.  Find the slope of the line from (36, 49.5) to (57, 51.5).

m = (51.5 − 49.5) / (57 − 36)

m = 2/21

So at x = 51:

2/21 = (y − 49.5) / (51 − 36)

y = 50.93

Answer:

Step-by-step explanation:

The determinant of the coefficient matrix is ...

  [tex]D=\left|\begin{array}{ccc}2&5&3\\4&-1&-4\\-5&-2&6\end{array}\right|\\\\=2(-1)(6)+5(-4)(-5)+3(4)(-2)-2(-4)(-2)-5(4)(6)-3(-1)(-5)\\\\=-12+100-24-16-120-15=\boxed{-87}[/tex]

The other determinants are found in similar fashion after substituting the constants on the right for each of the above matrix columns, in turn.

Those determinants are ...

  [tex]D_x=\left|\begin{array}{ccc}21&5&3\\-13&-1&-4\\0&-2&6\end{array}\right|=174[/tex]

  [tex]D_y=\left|\begin{array}{ccc}2&21&3\\4&-13&-4\\-5&0&6\end{array}\right|=-435[/tex]

  [tex]D_z=\left|\begin{array}{ccc}2&5&21\\4&-1&-13\\-5&-2&0\end{array}\right|=0[/tex]

The solutions are ...

  x = 174/-87 = -2

  y = -435/-87 = 5

  z = 0

That is, (x, y, z) = (-2, 5, 0).

Answer:

A) Reject H0 if F > 5.417

B) we fail to reject the null hypothesis and conclude that we do not have sufficient evidence at 0.05 level of significance to support the claim that there is a difference in the mean number of months before a raise was granted among the four CPA firms

Step-by-step explanation:

A) From the table, we can see that we have df1 = 3 and df2 = 15. And we are given a significance level of α = 0.01

We are also given f-value of 1.75

Thus,from the f-distribution table attached at significance level of α = 0.01 and df1 = 3 and df2 = 15, we have;

F-critical = 5.417

Normally, we reject H0 if F > 5.417

But in this case, F is 1.75 < 5.417 and so we conclude that we do not reject H0 at the 0.01 level of significance

B) for 0.05 level of significance, df1 = 3 and df2 = 15, from the 2nd table attached, we have;

F-critical = 3.2874

Again the f-value is less than this critical one.

Thus, we fail to reject the null hypothesis and conclude that we do not have sufficient evidence at 0.05 level of significance to support the claim that there is a difference in the mean number of months before a raise was granted among the four CPA firms

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

∴

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

∴

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

Answer:

It sold 14 cans boxes of food and 12 cans of food.

Step-by-step explanation:

The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.

Let the no. of sets of food boxes be x.

According to the question,

6x+7x=26

13x=26

x=26/13

x=2

No. of food cans =6x=6×2=12 cans

No. of food boxes=7x=7×2=14 boxes

Please mark brainliest ,if it is truly the best ! Thank you!

Answer:

1/3(a+b)

hope it helps :>

Answer:

The answer is D)90

Hope I helped

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

Answer:

[tex]\mathbf{P(X=5) =0.0888}[/tex]    

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

[tex]\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]

[tex]\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]

where;

n = 8 and π = 0.36

For x = 5

The probability [tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}[/tex]

[tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}[/tex]

[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}[/tex]

[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}[/tex]

[tex]\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}[/tex]

[tex]\mathtt{P(X=5) =0.0887645}[/tex]

[tex]\mathbf{P(X=5) =0.0888}[/tex]     to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5)[tex]\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})[/tex]

[tex]{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +[/tex][tex]\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )[/tex]

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1  - P( x  ≤ 5 )

P ( x ≥ 6) = 1  - 0.9707

P ( x ≥ 6) = 0.0293

Answer:

Perimeter = 42 cm

Step-by-step explanation:

A square has all equal sides so you would just add 10.5 + 10.5 + 10.5 + 10.5 to get 42 cm.

Answer:

42 cm

Step-by-step explanation:

Side of square = 10.5 cm (given)

Perimeter of square = Side X 4

                                  = 10.5 X 4

                                  = 42 cm

HOPE THIS HELPED YOU !

:)

Answer:

Three to the second power

Step-by-step explanation:

Hey there!

3 square

Can be written as the following,

Three to the second power

Hope this helps :)

Answer:

eddfdgdccggģdffcdrrfxddxcvgfx

Complete Question

The complete question is shown on the first uploaded image

Answer:

Option C is the correct option

Step-by-step explanation:

From the question we are told that

   The equation is  [tex]f (x, y , z ) = x^2 +y^2 + z^2[/tex]

    The constraint is  [tex]P(x, y , z) = x + y + z - 24 = 0[/tex]

Now using Lagrange multipliers  we have that  

   [tex]\lambda = \frac{ \delta f }{ \delta x } = 2 x[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta y } = y[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta z } = 2 z[/tex]

=>       [tex]x = \frac{ \lambda }{2}[/tex]

          [tex]y = \frac{ \lambda }{2}[/tex]

         [tex]z = \frac{ \lambda }{2}[/tex]

From the constraint  we have

      [tex]\frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{2} = 24[/tex]

=>   [tex]\frac{3 \lambda }{2} = 24[/tex]

=>   [tex]\lambda = 16[/tex]

substituting for x, y, z

=>   x =  8

=>  y =  8

=>   z =  8        

Hence

    [tex]f (8, 8 , 8 ) = 8^2 +8^2 + 8^2[/tex]

    [tex]f (8, 8 , 8 ) = 192[/tex]

Answer:

Step-by-step explanation:

This is best read on the number line.

Look at the picture.

[tex]-\dfrac{17}{2}=-8\dfrac{1}{2}=-8.5[/tex]

Answer:

Hey there!

You can think of the rate of change as the slope of a quadratic function- here we see that it is 9/-3, or - 3.

Let  me know if this helps :)

Answer:

–3 meters per second

Step-by-step explanation:

Answer:

The probability of raining if the number of days is more than 23 is 0.0668.

Step-by-step explanation:

We are given that the mean number of days to observe rain in a particular city is 20 days with a standard deviation of 2 days.

Let X = Number of days of observing rain in a particular city.

The z-score probability distribution for the normal distribution is given by;

                         Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean number of days = 20 days

           [tex]\sigma[/tex] = standard deviation = 2 days

So, X ~ Normal([tex]\mu=20, \sigma^{2} = 2^{2}[/tex])

Now, the probability of raining if the number of days is more than 23 is given by = P(X > 23 days)

        P(X > 23 days) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{23-20}{2}[/tex] ) = P(Z > 1.50) = 1 - P(Z [tex]\leq[/tex] 1.50)

                                                              = 1 - 0.9332 = 0.0668

The above probability is calculated by looking at the value of x = 1.50 in the z table which has an area of 0.9332.

it was all about equating some values

to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.

Let's determine the number of cooks required to bake the required number of cakes during the available time.

We have the following information:

- Each cook can bake either 5 large cakes or 14 small cakes per hour.

- The kitchen is available for 3 hours.

- We need to bake 29 large cakes and 260 cakes in total.

First, let's calculate the number of large cakes that can be baked by one cook in 3 hours:

1 cook can bake 5 large cakes/hour × 3 hours = 15 large cakes.

Next, let's calculate the number of small cakes that can be baked by one cook in 3 hours:

1 cook can bake 14 small cakes/hour × 3 hours = 42 small cakes.

Now, let's calculate the number of large cakes that can be baked by all the cooks in 3 hours:

Total number of large cakes = Number of cooks × Large cakes per cook per 3 hours

We need to bake 29 large cakes, so:

29 = Number of cooks × 15

Number of cooks = 29 / 15 ≈ 1.93

Since we can't have a fraction of a cook, we need to round up to the nearest whole number. Therefore, we need at least 2 cooks to bake the required number of large cakes.

Similarly, let's calculate the number of small cakes that can be baked by all the cooks in 3 hours:

Total number of small cakes = Number of cooks × Small cakes per cook per 3 hours

We need to bake 260 small cakes, so:

260 = Number of cooks × 42

Number of cooks = 260 / 42 ≈ 6.19

Again, rounding up to the nearest whole number, we need at least 7 cooks to bake the required number of small cakes.

Since we need to satisfy both requirements for large and small cakes, we choose the larger number of cooks required, which is 7 cooks.

Therefore, to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.

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Answer:

False

Step-by-step explanation:

You Cnat solve it

Answer:

you cannot solve it

Step-by-step explanation:

false

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]