The following is a scatterplot of the percent of children under age 18 who are not in school or in the labor force vs. the number of juvenile violent crime arrests for each of the 50 states. The least-squares regression line has been drawn in on the plot. We would like to predict what the number of juvenile violent crime arrests would be in a state if 25% of children are not in school or in the labor force. This is called

Answer:

hi, option C is correct because it has a right angel. please give brainliest

Answer:

3/20

Step-by-step explanation:

By question it's given that ,

[tex]\implies \dfrac{a}{b}=\dfrac{2}{5}[/tex]

[tex]\implies \dfrac{b}{c}=\dfrac{3}{8}[/tex]

And we need to find out the value of a/c .For that Multiply both of them , we have ;

[tex]\implies \dfrac{a}{b} \times\dfrac{b}{c}=\dfrac{2}{5}\times \dfrac{3}{8} \\\\\implies \dfrac{a}{c}= \dfrac{3}{20}[/tex]

Hence the required answer is 3/20 .

Step-by-step explanation:

Let's say the rectangle's width is equal to y. We know that the length is three times the width, so the length = 3 * y. We also know that the area for a rectangle is equal to length * width, so the area, z, is equal to

(3*y) * y = z

3 * y² = z

Now, let's increase the width of the rectangle by 1. We can replace y with y+1 (as y+1 is 1 greater than y), and 3 * y with 3 * (y+1) to get

3*(y+1) * (y+1) = new area

(3y+3)*(y+1) = new area

3y²+3y +3 y + 3 = new area

3y² + 6y + 3 = new area

The difference in area is equal to the new area subtracted by the old area, or

3y²+6y+3 - 3y² = 6y +3. The variable for x is not given, so if x = (2y+1), the answer would be the second choice. However, solely using the information given, it is impossible to determine a solution outside of saying that it is not option 4, as 6y + 3 ≠ 3

Answer:

0.417

Step-by-step explanation:

Just divide 12/5 and the answer is 0.416666666...

Round up and you get 0.417.

Hope this helped!

Answer:

0.663 ( c )

Step-by-step explanation:

U1 = 0.1 , U2 = 0.8

using the "cosine" version of Box-Muller to generate a single Nor(-1,4) random variable

first step : generate single obsⁿ from N ( -1,4 )

attached below is the detailed solution

Answer:

h(t) = -16t(t-6)

h(2) = 128

Step-by-step explanation:

h(t) = -16t² + 96t

h(t) = -16t(t-6)

t = 3

h(2) = -16(2)(2 - 6)

h(2) = 128

Answer:

24

Step-by-step explanation:

You first find out how many cubes can fit into each measurement, then multiply them. (2*4*3=24)

Answer:

It will take 24 cubes to fill the rectangular prism.

Step-by-step explanation:

Find the volume of a cube with side lengths of 1/2 cm:

1/2^3 = 1/8

1/8 cm^3

Find the volume of the whole rectangular prism (lwh):

1 x 3/2 x 2

= 3/2 x 2

= 3

3 cm^3

Divide the volume of the prism by the bolume of one cube:

3 ÷ 1/8 = 24

Therefore it will take 24 cubes to fill the prism. Hope this helps!

Answer:

35x^2 - 28x - 7

Step-by-step explanation:

Answer:

x = 2

x =-2

x =1 +√3i

x =1 -√3i

x =-1 +√3i

x =-1 -√3i  

Step-by-step explanation:

[tex]x^{6} -64 = 0\\(x^{3} -8)(x^{3} + 8) = 0\\ \\( x - 2) (x^{2} + 2x + 4)( x + 2) (x^{2} -2x + 4) = 0\\[/tex]

x = 2

x =-2

x =1 +√3i

x =1 -√3i

x =-1 +√3i

x =-1 -√3i  

16:52

Hope this helps! :)

Step-by-step explanation:

ejejejejrjruruehehhr

Answer:

Step-by-step explanation:

Um where is the diagrahm

Answer:

9a²b

Step-by-step explanation:

Hi there!

We need to find the greatest common factor out of 45a²b³ and 18[tex]a^{4}[/tex]b

We can split apart the monomials to make it easier

45a²b³ is 45*a²b³

18[tex]a^{4}[/tex]b is 18*[tex]a^{4}[/tex]b

First, let's find the GCF out of 45 and 18 (the number coefficients)

we can find all of the multiples of the 2 numbers:

45 is made up of 9 and 5

9 is made up of 3 and 3

so 3*3*5 is 45

18 is made up of 2 and 9

9 is made up of 3 and 3

so 2*3*3 is 18

3*3 is in both 45 and 18, so 9 is the GCF out of 45 and 18

Now let's find the GCF out of a²b³ and [tex]a^{4}[/tex]b

a²b³ made up of a² and b³

so a²b³ is a*a*b*b*b

[tex]a^{4}[/tex]b is made up of [tex]a^{4}[/tex] and b

so [tex]a^{4}[/tex]b is a*a*a*a*b

a*a*b is in both a²b³ and [tex]a^{4}[/tex]b, so the GCF out of a²b³ and [tex]a^{4}[/tex]b is a²b

Now multiply 9 and a²b together, as they are only the GCF of the parts of the monomials

9*a²b=9a²b

there's the greatest common factor of the 2 monomials

Hope this helps!

Answer:

2 Answers. #1. +11. [3+(3/7)] times 7 is 24. DarkBlaze347 May 1, 2015. +5. Good job, DB! civonamzuk May 1, 2015.

35 Online Users.

Step-by-step explanation:

brainliest please and follow:D

Answer: The answer is 2,4. May I have the brainiest? pls I only need one more.

Answer:

36281629273781646181993836619946527189119292937467482919198$7473828191927364732818919283838292927383883829118661552621718919191019284746617171819001187373765252728

Step-by-step explanation:

173899918377+28910873638282

hiiksbsjxbxjsoahwjsissnsks

Answer:

About 4 minutes and 3 seconds

Step-by-step explanation:

If I have 55 invitations to be created in four hours.

4 hours × 60 =240 minutes

240/55= 4.36 minutes

So if I spend 4 minutes and 3 seconds on an invitation, I should be able to fill the order.

Answer:

2.2763 x 10 to the power of 4

for some reason it doesn't let me put in the explanation

Answer:

Step-by-step explanation:

Given expressions are,

[tex]p=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}+\sqrt{5}}[/tex] and [tex]q=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]

Remove the radicals from the denominator from both the expressions.

[tex]p=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}+\sqrt{5}} \times \frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]

  [tex]=\frac{(\sqrt{10}-\sqrt{5})^2}{(\sqrt{10})^2-(\sqrt{5})^2}[/tex]

  [tex]=\frac{(\sqrt{10}-\sqrt{5})^2}{5}[/tex]

[tex]\sqrt{p}=\sqrt{\frac{(\sqrt{10}-\sqrt{5})^2}{5}}[/tex]

      [tex]=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{5}}[/tex]

[tex]q=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]

  [tex]=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}\times \frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}+\sqrt{5}}[/tex]

  [tex]=\frac{(\sqrt{10}+\sqrt{5})^2}{(\sqrt{10})^2-(\sqrt{5})^2}[/tex]

  [tex]=\frac{(\sqrt{10}+\sqrt{5})^2}{5}[/tex]

[tex]\sqrt{q}=\sqrt{\frac{(\sqrt{10}+\sqrt{5})^2}{5}}[/tex]

     [tex]=\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}}[/tex]

[tex]\sqrt{q}-\sqrt{p}-2\sqrt{pq}=\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}}-\frac{(\sqrt{10}-\sqrt{5})}{\sqrt{5}}-2(\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}})(\frac{(\sqrt{10}-\sqrt{5})}{\sqrt{5}})[/tex]

                           [tex]=\frac{1}{\sqrt{5}}(\sqrt{10}+\sqrt{5}-\sqrt{10}+\sqrt{5})-\frac{2}{5}[(\sqrt{10})^2-(\sqrt{5})^2)][/tex]

                           [tex]=\frac{1}{\sqrt{5}}(2\sqrt{5})-\frac{2}{5}(10-5)[/tex]

                           [tex]=2-2[/tex]

                           [tex]=0[/tex]

Answer:

Tenemos dos propiedades de la potencia en este caso:

Para un numero real A:

[tex]A^0 = 1[/tex]

[tex](A^n)^m = A^{n*m}[/tex]

En este caso nuestra ecuación es:

[tex][ [(\frac{0.1234}{-3.2098})^4]^3]^0[/tex]

usando la segunda propiedad, podemos reescribir como:

[tex][ [(\frac{0.1234}{-3.2098})^4]^3]^0 = (\frac{0.1234}{-3.2098})^{4*3*0} = (\frac{0.1234}{-3.2098})^0[/tex]

Y acá tenemos un numero real a la potencia 0, sabemos que esto es igual a 1, entonces:

[tex](\frac{0.1234}{-3.2098})^0 = 1[/tex]

Answer:

[tex]P(3) = ^9C_3 * 0.5^3 *0.5^6[/tex]

Step-by-step explanation:

Given

[tex]n = 9[/tex] --- number of flips

Required

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

The probability of getting a head is:

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

[tex]p = 0.5[/tex]

The distribution follows binomial probability, and it is calculated using:

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

So, we have:

[tex]P(3) = ^9C_3 * 0.5^3 * (1 - 0.5)^{9-3}[/tex]

[tex]P(3) = ^9C_3 * 0.5^3 *0.5^6[/tex]

Answer:

Aron flips a penny 9 times. Which expression represents the probability of getting exactly 3 heads?

Answer: A

Step-by-step explanation:

Answer:

-3072

Step-by-step explanation:

Answer:

[tex]4 {x}^{2} - 4x - 8 - (2 {x}^{2} + 3x - 6) = 4 {x}^{2} - 4x - 8 - 2 {x}^{2} - 3x + 6 = 2 {x}^{2} - 7x - 2[/tex]

Answer:

a) By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.

b) Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.

c) 0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less

d) 0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean 0.9 and standard deviation 1.1.

This means that [tex]\mu = 0.9, \sigma = 1.1[/tex]

Suppose a television network selects a random sample of 1000 families in the United States for a survey on TV viewing habits.

This means that [tex]n = 1000, s = \frac{1.1}{\sqrt{1000}} = 0.035[/tex]

a. Describe (as shape, center and spread) the sampling distribution of the possible values of the average number of children per family.

By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.

b. What average numbers of children are reasonably likely in the sample?

By the Empirical Rule, 95% of the sample is within 2 standard deviations of the mean, so:

0.9 - 2*0.035 = 0.83

0.9 + 2*0.035 =  0.97

Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.

c. What is the probability that the average number of children per family in the sample will be 0.8 or less?

This is the p-value of Z when X = 0.8. So

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

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]

[tex]Z = -2.86[/tex]

[tex]Z = -2.86[/tex] has a p-value of 0.0021

0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less.

d. What is the probability that the average number of children per family in the sample will be between 0.8 and 1.0?

p-value of Z when X = 1 subtracted by the p-value of Z when X = 0.8.

X = 1

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{1 - 0.9}{0.035}[/tex]

[tex]Z = 2.86[/tex]

[tex]Z = 2.86[/tex] has a p-value of 0.9979

X = 0.8

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]

[tex]Z = -2.86[/tex]

[tex]Z = -2.86[/tex] has a p-value of 0.0021

0.9979 - 0.0021 = 0.9958

0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0

[tex] \sqrt[5]{32} \times 2 \sqrt[3]{81} \\ = {32}^{ \frac{1}{5} } \times 2 {(81)}^{ \frac{1}{3} } \\ = ({{2}^{5}})^{ \frac{1}{5} } \times {2({3}^{3})}^{ \frac{1}{3} } \\ = {2}^{1} \times 2({3}^{1}) \\ = 2 \times 2 \times 3 \\ = 4 \times 3 \\ = 12[/tex]

This is the answer.

Hope it helps!!