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3 years ago

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Jill aunt got her a $50 gift card to her favorite boutique she has already picked out 4 shirts that cost $8.00 each how many pai

rs of shorts can she buy if shorts cost $10 a pair?

1 answer:

kenny6666 [7]3 years ago

6 0

Answer:

1

Step-by-step explanation:

If she has four $8 shirts, that costs $32, leaving her with $18 dollars, only enough for 1 pair of $10 shorts

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Someone please help me?!?!

VikaD [51]

Answer:

30 is correct answer. ... ...........

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3 years ago

Which equation(s) represents a proportional relationship? A) y = 5+x B) y = 4x C) y = 7x? D) y = 2-x

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Answer:

B & C

Step-by-step explanation:

A proportional relationship has a proportionality constant (constant rate of change) and and must cross through the origin (or begin at (0,0). This means it's y-intercept is 0. Each of the options is a linear function. All of the functions have a constant rate of change. Only one as b=0 or 0 as the y-intercept.

B) y=4x has a constant of 4 and passes through the origin.

C) y=7x has a constant of 7 and passes through the origin.

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3 years ago

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4 years ago

Read 2 more answers

Which figures have a volume greater than 500 cubic inches? Choose all that apply.

yaroslaw [1]

Answer:

All shapes except the cone with 5in radius and 8in height

Step-by-step explanation:

Volume of the cylinder with radius 5in and height 8in is expressed as;

V = πr²h

V = 3.14*5²*8

V = 3.14*25*8

V = 628 square inches

The volume of the cylinder is greater than 500in³

Volume of the cone with radius 5in and height 24in is expressed as;

V = 1/3πr²h

V = 1/3*3.14*5²*24

V = 3.14*25*8

V = 628 square inches

The volume of the cone is greater than 500in³

Volume of the spheree with radius 5in is expressed as;

V = 4/3πr³

V = 4/3π(5)³

V = 4/3 * 3.14 * 125

V = 523.33 square inches

The volume of the sphere is greater than 500in³

6 0

3 years ago

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

frosja888 [35]

Answer:

a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

b) 0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday

d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

Step-by-step explanation:

We solve this question using the normal approximation to the binomial distribution.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

Sample of 723, 3.7% will live past their 90th birthday.

This means that $n = 723, p = 0.037$ .

So for the approximation, we will have:

$\mu = E(X) = np = 723*0.037 = 26.751$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{723*0.037*0.963} = 5.08$

(a) 15 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 15 - 0.5) = P(X \geq 14.5)$ , which is 1 subtracted by the pvalue of Z when X = 14.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{14.5 - 26.751}{5.08}$

$Z = -2.41$

$Z = -2.41$ has a pvalue of 0.0080

1 - 0.0080 = 0.9920

0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 30 - 0.5) = P(X \geq 29.5)$ , which is 1 subtracted by the pvalue of Z when X = 29.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{29.5 - 26.751}{5.08}$

$Z = 0.54$

$Z = 0.54$ has a pvalue of 0.7054

1 - 0.7054 = 0.2946

0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

This is, using continuity correction, $P(25 - 0.5 \leq X \leq 35 + 0.5) = P(X 24.5 \leq X \leq 35.5)$ , which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So

X = 35.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{35.5 - 26.751}{5.08}$

$Z = 1.72$

$Z = 1.72$ has a pvalue of 0.9573

X = 24.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{24.5 - 26.751}{5.08}$

$Z = -0.44$

$Z = -0.44$ has a pvalue of 0.3300

0.9573 - 0.3300 = 0.6273

0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday.

(d) more than 40 will live beyond their 90th birthday

This is, using continuity correction, P(X > 40+0.5) = P(X > 40.5), which is 1 subtracted by the pvalue of Z when X = 40.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40.5 - 26.751}{5.08}$

$Z = 2.71$

$Z = 2.71$ has a pvalue of 0.9966

1 - 0.9966 = 0.0034

0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

6 0

3 years ago