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stealth61 [152]

3 years ago

11

Callie solved this one-variable equation. When she was done, she tried to verify her answer, but found that it was wrong. Can yo

u find her mistake? 1 2 (4k + 8) = 10 1. 2k + 16 = 10 2. 2k = −6 3. k = −3

2 answers:

Naddik [55]3 years ago

5 0

Answer:

Mistake found in step 1

Step-by-step explanation:

Callie solved this one-variable equation. When she was done, she tried to verify her answer, but found that it was wrong.

$2(4k + 8) = 10$

Step 1: $2k + 16 = 10$

Step 2: $2k = -6$

Step 3: $k = -3$

In step 2, 2 is distributed inside the parenthesis

$2(4k + 8) = 10$

When 2 is multiplied inside the parenthesis, the equation becomes

$8k + 16 = 10$

She forgot to multiply 2 with 4k

So the mistake is in step 1

Andrews [41]3 years ago

3 0

2(4k+8)=10
8k+16=10 (she said 2k instead of 8k)

To continue solving, subtract 16 from both sides. That would give you 8k= -6. Then divide both sides by 8, and get -6/8 which equals -3/4.

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2 years ago

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At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro

Gnom [1K]

Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of $\pi$ , so we use the worst case scenario, which is $\pi = 0.5$

Then

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}$

$n = 1067.11$

Rounding up

A sample of 1068 is needed.

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3. From the table below, find Prof. Xin expected value of lateness. (5 points) Lateness P(Lateness) On Time 4/5 1 Hour Late 1/10

wariber [46]

Answer:

The expected value of lateness $\frac{7}{20}$ hours.

Step-by-step explanation:

The probability distribution of lateness is as follows:

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2 Hours Late 1/20

3 Hours Late 1/20

The formula of expected value of a random variable is:

$E(X)=\sum x\cdot P(X=x)$

Compute the expected value of lateness as follows:

$E(X)=\sum x\cdot P(X=x)$

$=(0\times \frac{4}{5})+(1\times \frac{1}{10})+(2\times \frac{1}{20})+(3\times \frac{1}{20})\\\\=0+\frac{1}{10}+\frac{1}{10}+\frac{3}{20}\\\\=\frac{2+2+3}{20}\\\\=\frac{7}{20}$

Thus, the expected value of lateness $\frac{7}{20}$ hours.

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