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3 years ago

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Problem 2 – SHSAT Question Lily spent one-fifth of her salary on a pair of new shoes and then received $90 as a birthday present

from her mom. If Lily now has $340, how much money did she begin with?
A. $360
B. $340
C. $312.50
D. $280

1 answer:

const2013 [10]3 years ago

6 0

Answer: $312.5

Step-by-step explanation:

Lily spent one-fifth of her salary on a pair of shoes and then got $90 as a birthday present from her mom and was left with $340. To calculate the amount Lily had at first goes thus:

Since Lily spent 1/5 on shoes, that means she had (1 - 1/5) = 4/5 left.

Let the amount she had at first be y.

(4/5 of y) + 90 = 340

(4/5 × y) + 90 = 340

0.8y + 90 = 340

0.8y = 340 - 90

0.8y = 250

Divide both side by 0.8

0.8y/0.8 = 250/0.8

y= 312.5

She had $312.5 at first.

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A survey sampled men and women workers and asked if they expected to get a raise or promotion this year. Suppose the survey samp

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Answer:

a)

The null hypothesis is: $H_0: p_M - p_W = 0$

The alternative hypothesis is: $H_1: p_M - p_W > 0$

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c) The p-value of the test is 0.0039 < 0.01, which means that the results are statistically significant so that you can conclude a greater proportion of men expect to get a raise or a promotion this year.

Step-by-step explanation:

Before solving this question, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

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

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

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Men:

98 out of 200, so:

$p_M = \frac{98}{200} = 0.49$

$s_M = \sqrt{\frac{0.49*0.51}{200}} = 0.0353$

Women:

72 out of 200, so:

$p_W = \frac{72}{200} = 0.36$

$s_W = \sqrt{\frac{0.36*0.64}{200}} = 0.0339$

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At the null hypothesis, we test if the proportion are similar, that is, if the subtraction of the proportions is 0, so:

$H_0: p_M - p_W = 0$

At the alternative hypothesis, we test if the proportion of men is greater, that is, the subtraction is greater than 0, so:

$H_1: p_M - p_W > 0$

b. What is the sample proportion for men? For women?

For men is of 0.49 and for women is of 0.36.

c. Use α= 0.01 level of significance. What is the p-value and what is your conclusion?

From the sample, we have that:

$X = p_M - p_W = 0.49 - 0.36 = 0.13$

$s = \sqrt{s_M^2+s_W^2} = \sqrt{0.0353^2 + 0.0339^2} = 0.0489$

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, and s is the standard error, so:

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The p-value of the test is the probability of finding a difference above 0.13, which is the p-value of z = 2.66.

Looking at the z-table, z = 2.66 has a p-value of 0.9961.

1 - 0.9961 = 0.0039.

The p-value of the test is 0.0039 < 0.01, which means that the results are statistically significant so that you can conclude a greater proportion of men expect to get a raise or a promotion this year.

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Step-by-step explanation:

We need to find value of x in the equation: $2^{2x}+ 7 = 2^{15}$

This can be solved using natural log

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Subtract 7 from both sides:

$2^{2x}+ 7-7 = 2^{15}-7$

Simplifying:

$2^{2x}=32768-7$

$2^{2x}=32761$

Using rule if f(x)=g(x) then, ln(f(x))=ln(g(x))

So,

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Applying log rule: logₐ(x^b)= b logₐ(x)

$2x(ln(2))=ln(32761)$

Divide both sides by 2ln(2)

$2x(ln(2))=[tex]\frac{2x(ln(2))}{2(ln(2))}=\frac{ln(32761)}{2(ln(2))} \\x=\frac{ln(32761)}{2(ln(2))}\\Rewriting\,\,32761\,\,as 181^2\\x=\frac{ln(181^2)}{2(ln(2))}\\x=\frac{2ln(181)}{2(ln(2))}\\x=\frac{ln(181)}{ln(2)}\\x=7.499$

So, Value of x = 7.499

Keywords: Solving Equations

Learn more about Solving Equations at:

#learnwithBrainly

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Step-by-step explanation:

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