4 + {−5 + [−3 + 4 + 2(−7 + 4) + 4] + 2}
7 + {−2 + [5 + 4(−3) + (−6 + 2) + (−3)]}
(−2) + {3 + [−4 + (3)] + 7} + (−5)
(−10) + {−7 + [−4 + 5(−9) + 5] + 8}
18 + 3{−4 + [−15 + 20 + (−3)] + (−9)}
−(−8 + 5) + {−4 + [−7 + (−9 − 5) + 3]}
− 9 + {−7 − 2[−(4 + 1) + (5 − 9)]} − 3
−1 − 3{6 − [4 + 2(−7 + 8) − 5] − 2}
−(−2) − {−(−7) + [−3 + (−5 − 2) + 6]}
−(−6 − 2) − 2{−4(−5) + 5[−3 + (−2)]}
6 − 2{−4 + 3[3 − 3(−8 − 5) + 14] − 7}
Me pueden ayudar por favor es urgente!!! y Buenas Noches ;)

Answer:

2, 3, 7 or 8

Step-by-step explanation:

Answer:

B. x=4

Step-by-step explanation:

I hope this helps!

Answer:

See Explanation

Step-by-step explanation:

Given

[tex]0 \to[/tex] Correct

[tex]1 \to[/tex] Defective

Required

The probability that the next is defective

The question is incomplete because the list of bottles that came out is not given.

However, the formula to use is:

[tex]Pr = Num ber\ o f\ d e f e c t i v e \div T o t a l\ b o t t l e s[/tex]

Take for instance, the following outcomes:

[tex]0\ 1\ 0\ 0\ 0\ 1\ 0\ 0\ 1\ 1\ 0\ 0\ 0\ 1[/tex]

We have:

[tex]Total = 14[/tex]

[tex]D e f e ctive = 9[/tex] --- i.e. the number of 0's

So, the probability is:

[tex]Pr = 9 \div 14[/tex]

[tex]Pr = 0.643[/tex]

Answer:

1/20

Step-by-step explanation:

000000000000100000

It is asking the probability of a defective bottle

there is one defective bottle out of 20

Step-by-step explanation:

4² = 16

5² = 25

16+25 = 41

41 is the answer.

Hope it helps! :)

Answer:

[tex]( {4}^{2} + {5}^{2} ) \\ (16 + 25) \\ = 41[/tex]

Answer:

90

Step-by-step explanation:

Step-by-step explanation:

A

=

h

b

b

2

=

11

·

22

2

=

121

cm²

The area of the obtuse triangle with a height of 11 cm and a base of 22 cm is 121 cm².

To determine the area of an obtuse triangle, we can use the formula for the area of a triangle, which is given by:

Area = (1/2) * base * height

In this case, the height of the triangle is given as 11 cm and the base is given as 22 cm.

Substituting these values into the formula, we have:

Area = (1/2) * 22 cm * 11 cm

Calculating this expression, we get:

Area = (1/2) * 242 cm²

Simplifying further, we have:

Area = 121 cm²

The area of a triangle is calculated by multiplying half of the base by the height. In this case, the given height is 11 cm and the base is 22 cm. Substituting these values into the formula, the area of the obtuse triangle is calculated to be 121 cm².

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Answer:

9 possible values of length and breadth.

Step-by-step explanation:

Given that the area of the rectangle is 18cm² . And the lengths of the sides are whole numbers. We need to tell the possible dimensions of the rectangle .

Let :-

[tex]\rm\implies Length = x [/tex]

[tex]\rm\implies Breadth = y [/tex]

We know that :-

[tex]\rm\implies Area = length \times breadth [/tex]

[tex]\rm\implies Area = x y \\\\\rm\implies xy = 18cm^2 [/tex]

Now look out for the possible factors of 18 . That is , ( 1 × 18 ) , ( 2 × 9) , ( 3 × 6 ) , ( 6 × 3 ) , ( 9 × 2 ) , (18 × 1 ) . That is total 9 possible values .

Answer:

18 pencils in 3 packs.

Step-by-step explanation:

Assuming that each pack will have the same amount of pencils. It is given that there are 36 pencils in all when one has 6 packs. Find the amount in each pack by dividing (total amount of pencils)/(amount of packs) = Amount of pencils per pack:

36/6 = 6

There are 6 pencils in each pack.

Now, Elsa wants to know how much pencils are in 3 packs. Multiply the amount of packs with the amount of pencils in each pack:

Total amount of pencils = amount of packs (3) x amount of pencils per pack (6)

                                       = 3 x 6

                                       = 18

There are 18 pencils in 3 packs.

~

Given:

AB formed by (-2,13) and (0,3).

CD formed by (-5,0) and (10,3).

To find:

Whether the segments AB and CD are parallel, perpendicular, or neither.

Solution:

Slope formula:

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

AB formed by (-2,13) and (0,3). So, the slope of AB is:

[tex]m_1=\dfrac{3-13}{0-(-2)}[/tex]

[tex]m_1=\dfrac{-10}{2}[/tex]

[tex]m_1=-5[/tex]

CD formed by (-5,0) and (10,3). So, slope of CD is:

[tex]m_2=\dfrac{3-0}{10-(-5)}[/tex]

[tex]m_2=\dfrac{3}{10+5}[/tex]

[tex]m_2=\dfrac{3}{15}[/tex]

[tex]m_2=\dfrac{1}{5}[/tex]

Since [tex]m_1\neq m_2[/tex], therefore the segments AB and CD are not parallel.

[tex]m_1\times m_2=-5\times \dfrac{1}{5}[/tex]

[tex]m_1\times m_2=-1[/tex]

Since [tex]m_1\times m_2=-1[/tex], therefore the segments AB and CD are perpendicular because product of slopes of two perpendicular lines is always -1.

Hence, the segments AB and CD are perpendicular.

Answer:

AB is perpendicular to CD.

Step-by-step explanation:

AB formed by (-2, 13) and (0, 3)

CD formed by (-5, 0) and (10, 3)

Slope of a line passing through two points is

[tex]m= \frac{y''-y'}{x''- x'}[/tex]

The slope of line AB is

[tex]m= \frac{3- 13}{0+2} = -5[/tex]

The slope of line CD is

[tex]m'= \frac{3 -0 }{10+5} = \frac{1}{5}[/tex]

As the product of m and m' is -1 so the lines AB and CD are perpendicular to each other.

Answer:

[tex]x = 0.143[/tex]

[tex]P(High\ |\ Cancer) = 0.215[/tex]

Not independent

Step-by-step explanation:

Given

See attachment for proper table

Solving (a): The missing probability

First, we add up the given probabilities

[tex]Sum = 0.242+0.047+0.079+0.391+0.098[/tex]

[tex]Sum = 0.857[/tex]

The total probability must add up to 1.

If the missing probability is x, then:

[tex]x + 0.857 = 1[/tex]

Collect like terms

[tex]x = -0.857 + 1[/tex]

[tex]x = 0.143[/tex]

Solving (b): P(High | Cancer)

This is calculated as:

[tex]P(High\ |\ Cancer) = \frac{n(High\ n\ Cancer)}{n(Cancer)}[/tex]

So, we have:

[tex]P(High\ |\ Cancer) = \frac{0.079}{0.242+0.047+0.079}[/tex]

[tex]P(High\ |\ Cancer) = \frac{0.079}{0.368}[/tex]

[tex]P(High\ |\ Cancer) = 0.215[/tex]

Solving (c): P(Leukemia) independent of P(High Wiring)

From the attached table

[tex]P(Leukemia\ n\ High\ Wiring) = 0.242[/tex]

[tex]P(Leukemia) = 0.242 + 0.391 =0.633[/tex]

[tex]P(High\ Wiring) = 0.242+0.047+0.079=0.368[/tex]

If both events are independent, then:

[tex]P(Leukemia\ n\ High\ Wiring) = P(Leukemia) * P(High\ Wiring)[/tex]

[tex]0.242 = 0.633 * 0.368[/tex]

[tex]0.242 \ne 0.232[/tex]

Since the above is an inequality, then the events are not independent

Answer:

17

Step-by-step explanation:

Answer:

240

Step-by-step explanation:

well do *

so 8x6x5 = 240 there's your answer

Answer:

[tex]S.A=1/2(8+8)(9^{2})+8\times 6+8\times 5[/tex]

[tex]=26\times2+48+40[/tex]

[tex]=140 ~cm^{2}[/tex]

-------------------------

HOPE IT HELPS

HAVE A GREAT DAY!!

X is a vertical angle to the angle marked as 100 degrees.

Vertical angles are the same so x = 100 degrees

Answer: 100 degrees

Answer:

Step-by-step explanation:   7  1  16   37

8/1     8 divided by 1

16/2   16 divided by 2

24/3   24 divided by 3       I could go on, but won't

Answer:

[tex]\begin{array}{cccccc}x & {0} & {1} & {2} & {3} & {4} \ \\ P(x) & {0.0256} & {0.1536} & {0.3456} & {0.3456} & {0.1296} \ \end{array}[/tex]

Step-by-step explanation:

Given

[tex]p = 60\%[/tex]

[tex]n = 4[/tex]

Required

The distribution of x

The above is an illustration of binomial theorem where:

[tex]P(x) = ^nC_x * p^x *(1 - p)^{n-x}[/tex]

This gives:

[tex]P(x) = ^4C_x * (60\%)^x *(1 - 60\%)^{n-x}[/tex]

Express percentage as decimal

[tex]P(x) = ^4C_x * (0.60)^x *(1 - 0.60)^{n-x}[/tex]

[tex]P(x) = ^4C_x * (0.60)^x *(0.40)^{4-x}[/tex]

When x = 0, we have:

[tex]P(x=0) = ^4C_0 * (0.60)^0 *(0.40)^{4-0}[/tex]

[tex]P(x=0) = 1 * 1 *(0.40)^4[/tex]

[tex]P(x=0) = 0.0256[/tex]

When x = 1

[tex]P(x=1) = ^4C_1 * (0.60)^1 *(0.40)^{4-1}[/tex]

[tex]P(x=1) = 4 * (0.60) *(0.40)^3[/tex]

[tex]P(x=1) = 0.1536[/tex]

When x = 2

[tex]P(x=2) = ^4C_2 * (0.60)^2 *(0.40)^{4-2}[/tex]

[tex]P(x=2) = 6 * (0.60)^2 *(0.40)^2[/tex]

[tex]P(x=2) = 0.3456[/tex]

When x = 3

[tex]P(x=3) = ^4C_3 * (0.60)^3 *(0.40)^{4-3}[/tex]

[tex]P(x=3) = 4 * (0.60)^3 *(0.40)[/tex]

[tex]P(x=3) = 0.3456[/tex]

When x = 4

[tex]P(x=4) = ^4C_4 * (0.60)^4 *(0.40)^{4-4}[/tex]

[tex]P(x=4) = 1 * (0.60)^4 *(0.40)^0[/tex]

[tex]P(x=4) = 0.1296[/tex]

So, the probability distribution is:

[tex]\begin{array}{cccccc}x & {0} & {1} & {2} & {3} & {4} \ \\ P(x) & {0.0256} & {0.1536} & {0.3456} & {0.3456} & {0.1296} \ \end{array}[/tex]

Answer:

i don't know how to work this

Answer:

x = [tex]\frac{3}{4}[/tex] , x = 20

Step-by-step explanation:

2([tex]\frac{3x-5}{x+2}[/tex] ) - 5 ([tex]\frac{x+2}{3x-5}[/tex] ) = 3

Multiply through by (x + 2)(3x - 5) to eliminate the fractions

2(3x - 5)² - 5(x + 2)² = 3(x + 2)(3x - 5) ← expand factors on both sides

2(9x² - 30x + 25) - 5(x² + 4x + 4) = 3(3x² + x - 10) ← distribute parenthesis

18x² - 60x + 50 - 5x² - 20x - 20 = 9x² + 3x - 30 ← simplify left side

13x² - 80x + 30  = 9x² + 3x - 30 ( subtract 9x² + 3x - 30 from both sides )

4x² - 83x + 60 = 0 factor by splitting the x- term

4x² - 80x - 3x + 60 = 0

4x(x - 20) - 3(x - 20) = 0

(4x - 3)(x - 20) = 0

Equate each factor to zero and solve for x

4x - 3 = 0 ⇒ 4x = 3 ⇒ x = [tex]\frac{3}{4}[/tex]

x - 20 = 0 ⇒ x = 20

Answer:

I believe its the 1st answer.

Answer:

The answer is y= - ⅓x + 5 in slope intercept form and y-7 = - ⅓ (x + 6) in point slope form.

Answer:

There is a non-removable discontinuity at x = 3

Step-by-step explanation:

We are  given that

[tex]f(x)=\frac{x^2+9}{x-3}[/tex]

We have to find true statement about given function.

[tex]\lim_{x\rightarrow 3}f(x)=\lim_{x\rightarrow 3}\frac{x^2+9}{x-3}[/tex]

=[tex]\infty[/tex]

It is not removable discontinuity.

x-3=0

x=3

The function f(x) is not define at x=3. Therefore, the function f(x) is continuous for all real numbers except x=3.

Therefore, x=3 is non- removable discontinuity of function f(x).

Hence, option B is correct.

[tex]\large\bold{\underline{\underline{ 4( {x - 7})^{2} }}}[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]

[tex]4x {}^{2} - 56x + 196[/tex]

Take [tex]4[/tex] as the common factor.

[tex] = 4( {x}^{2} - 14x + 49)[/tex]

[tex] = 4( {x}^{2} - 7x - 7x + 49)[/tex]

Taking [tex]x[/tex] as common from first two terms and [tex]7[/tex] from last two terms, we have

[tex] = 4 \: [ x(x - 7) - 7(x - 7) ][/tex]

Taking the factor [tex](x-7)[/tex] as common,

[tex] = 4(x - 7)(x - 7)[/tex]

[tex] = 4( {x - 7})^{2} [/tex]

[tex]\bold{ \green{ \star{ \red{Mystique35}}}}⋆[/tex]

Answer:

[tex]mDE=90[/tex]°

Step-by-step explanation:

In this problem, one is asked to find the measure of the arc in degrees. This problem provides one with a diagram since the center is unnamed, call the center point (O). One is given the information that angle (<DOE) has a measure of (90) degrees. One is asked to find the degree measure of the surrounding arc (DE).

The central angles theorem states that if an angle has its vertex on the center of a circle, the degree measure of the angle is equal to the degree measure of the surrounding arc. One can apply this here by stating the following:

[tex]m<DOE = mDE\\[/tex]

Substitute,

[tex]90=mDE[/tex]

Answer:

90g strawberry jelly

6 sponge fingers

315ml custard

135g tinned fruit

Step-by-step explanation:

----------------------------------------

Since those ingredients that are in the question say making a trifle for four people, let's divide each ingredient by four to find the ingredients for one person, and multiply by 3 to find the ingredients for 3 people.

So,

[tex]120/4=30[/tex]g strawberry jelly

[tex]8/4=2[/tex] sponge fingers

[tex]420/4=105[/tex]ml custard

[tex]180/4=45[/tex]g tinned fruit

There are the ingredients to make a trifle for one person.

-------------------->>>>

Now, let's multiply each of the ingredients for one person and multiply that by three to find the ingredients needed to make a trifle for four people.

[tex]30*3=90[/tex]g strawberry jelly

[tex]2*3=6[/tex] sponge fingers

[tex]105*3=315[/tex]ml custard

[tex]45*3=135[/tex]g tinned fruit

These are the ingredients needed to make a trifle for three people.

----------------------------------------

Hope this is helpful.

Answer:

True

Step-by-step explanation:

The equation of direct variation is

y = kx ← k is the constant of variation

To find k divide both sides by x

[tex]\frac{y}{x}[/tex] = k

That is the constant is the ratio of y- values to x- values

Given:

There are 35 times as many students at Wow University as teachers.

When all the students and teachers are seated in the 8544 seat auditorium, 12 seats are empty.

To find:

The total number of students.

Solution:

Let x be the number of teachers at Wow University. So, the number of student is :

[tex]35\times x=35x[/tex]

When all the students and teachers are seated in the 8544 seat auditorium, 12 seats are empty.

[tex]x+35x=8544-12[/tex]

[tex]36x=8532[/tex]

[tex]x=\dfrac{8532}{36}[/tex]

[tex]x=237[/tex]

The number of total students is:

[tex]35x=35(237)[/tex]

[tex]35x=8295[/tex]

Therefore, the total number of students is 8295.

Answer:

[tex]\frac{\sqrt{10} }{2}[/tex]

Step-by-step explanation:

on a 45 - 45 - 90 triangle if the sides are x then the hypotenuse is x[tex]\sqrt{2}[/tex]

so,

[tex]\sqrt{5}[/tex] = x[tex]\sqrt{2}[/tex]                  (Divide by [tex]\sqrt{2}[/tex] )

[tex]\frac{\sqrt{5} }{\sqrt{2} }[/tex] = x                        (rationalize the denominator)

[tex]\frac{\sqrt{5} }{\sqrt{2} }[/tex] · [tex]\frac{\sqrt{2} }{\sqrt{2} }[/tex] = [tex]\frac{\sqrt{10} }{\sqrt{4} }[/tex] = [tex]\frac{\sqrt{10} }{2}[/tex]

The length of side is √10/2

Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles.

The given triangle is a 45-90-45 so if the sides are x then the hypotenuse will be x√2

From the figure the hypotenuse of triangle is √5

SO equation it to x√2

We get √5=x√2

Divide both sides by √2

√5/√2=x

As the denominator should be a whole number we have to rationalize the denominator.

Multiply numerator and denominator with √2

√5/√2×√√2/√2

√10/2

Hence, the length of side is √10/2

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Answer:

Answer: The arc PS would be a minor arc · Step-by-step explanation: As a minor arc is one that is less that 180 degrees, it would be the only viable ...

Answer:

x = -0.70,-2.212.

Step-by-step explanation:

(a) The quadratic equation is :

[tex]6x^2 + 17x + 9 = 0[/tex]

Here,

a = 6, b = 17, c = 9

Put all the values in the quadratic formula.

[tex]x=\dfrac{-17+\sqrt{17^2-4\times 6\times 9}}{2(6)},\dfrac{-17-\sqrt{17^2-4\times 6\times 9}}{2(6)}\\\\x=-0.70,-2.12[/tex]

So, the solution of the given equation are -0.70,-2.212.

(b) Equation is : [tex]4x^2 -4x + 1 = 0[/tex].

Here, a = 4, b = -4 and c = 1

So,

[tex]x=\dfrac{-(-4)+\sqrt{(-4)^2-4\times 4\times 1}}{2(4)},\dfrac{-(-4)-\sqrt{(-4)^2-4\times 4\times 1}}{2(4)}\\\\x=0.5,0.5[/tex]

Hence, this is the required solution.

Answer:

£0.80

Step-by-step explanation:

1 box of eggs: £2.40

2 packs of bacon: 2 * £2.60

2 tins of baked beans: 2x

Change: 80p

Total = 2x + £2.40 + £5.20 + £0.80

Total = 2x + £8.40

2x + £8.40 = £10

2x = £1.60

x = £0.80

Answer: £0.80

Answer:

Profit obtained by X =  Rs. 2,976.64

Profit obtained by Y = Rs. 2,545.58

Profit obtained by Z = Rs. 2,777.78

Step-by-step explanation:

Total capital for the first 6 months = Rs 5000 + Rs.4500 + Rs.6500 = Rs. 16,000

Total capital for the next 3 months = Rs. 16,000+ Rs 5000 = Rs. 21,000

Total capital for the last 3 months of the year = Rs. 21,000 + (Rs 4500 * 2) = Rs. 30,000

Share of profit of each partner is the sum of all the ratios of his capital to total capital of the business at each point in time multiply by the ratio of the numbers of months covered by each capital to 12 months and then multiply by RS.8300.

Profit obtained by X = ((Rs 5000 / 16,000) * (6 / 12) * Rs. 8300) +  ((Rs 10,000 / 21,000) * (3 / 12) * Rs. 8300) + ((Rs 10,000 / 30,000) * (3 / 12) * Rs. 8300) =  Rs. 2,976.64

Profit obtained by Y = ((Rs 4500 / 16,000) * (6 / 12) * Rs. 8300) +  ((Rs 4500 / 21,000) * (3 / 12) * Rs. 8300) + ((Rs 13,500 / 30,000) * (3 / 12) * Rs. 8300) = Rs. 2,545.58

Profit obtained by Z = ((Rs 6500 / 16,000) * (6 / 12) * Rs. 8300) +  ((Rs 6500 / 21,000) * (3 / 12) * Rs. 8300) + ((Rs 6,500 / 30,000) * (3 / 12) * Rs. 8300) = Rs. 2,777.78

Confirmation of total profit shared = Rs. 2,976.64 + = Rs. 2,545.58 + Rs. 2,777.78 = Rs. 8,300