What is the inequality shown? n -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

Which of the following is equal to -5/4? A. -5/-4 B. -(-5/4) C. -(5/-4) D. -(-5/-4)

slavikrds [6]

Answer:

C. -(5/-4)

Step-by-step explanation:

$- \bigg(\frac{5}{ - 4} \bigg) \\ \\ = \frac{ - 5}{ - ( - 4)} \\ \\ = \frac{ - 5}{4} \\$

4 0

3 years ago

A stadium has 48,000 seats. Seats sell for $25 in Section A, $20 in Section B, and $15 in Section C. The number of seats in

Ira Lisetskai [31]

I had a super-ugly equation that didn’t work, but it got me close. (24,000/12,000/12,000 or so)
Then I adjusted the numbers to make the money work.

I’m sorry I can’t help more.

24,000(25) + 14,200(20) + 9,800(15)
= 600,000 + 284,000 + 147,000 = 1,031,000

Section A = 24,000 seats
Section B = 14,200 seats
Section C = 9,800 seats

14,200 + 9,800 = 24,000 seats

5 0

2 years ago

Mindy is the manager of the help desk at a large cable company. She notices that, on average, her staff can handle 50 calls/hr.

Hatshy [7]

Answer:

12

Step-by-step explanation:

6 0

2 years ago

Someoone plz help me

posledela

The ratio depicts that their is 63 boys and 16 girls based on the ratio 7:4.
I hope this helps!

7 0

2 years ago

The American Association of Individual Investors conducts a weekly survey of its members to measure the percent who are bullish,

vredina [299]

Answer:

1. There is not enough evidence to support the claim that bullish sentiment differs from its long-term average of 0.39.

2. There is enough evidence to support the claim that bearish sentiment is above its long-term average of 0.30.

Step-by-step explanation:

The question is incomplete:

The American Association of Individual Investors conducts a weekly survey of its members to measure the percent who are bullish, bearish, and neutral on the stock market for the next six months. For the week ending November 7, 2012 the survey results showed 38.5% bullish, 21.6% neutral, and 39.9% bearish (AAII website, November 12, 2012). Assume these results are based on a sample of 300 AAII members.

1. This is a hypothesis test for a proportion.

The claim is that bullish sentiment differs from its long-term average of 0.39.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.39\\\\H_a:\pi\neq 0.39$

The significance level is 0.05.

The sample has a size n=300.

The sample proportion is p=0.385.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.39*0.61}{300}}\\\\\\ \sigma_p=\sqrt{0.000793}=0.028$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.385-0.39+0.5/300}{0.028}=\dfrac{-0.003}{0.028}=-0.118$

This test is a two-tailed test, so the P-value for this test is calculated as:

$P-value=2\cdot P(z$

As the P-value (0.906) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that bullish sentiment differs from its long-term average of 0.39.

2) This is a hypothesis test for a proportion.

The claim is that bearish sentiment is above its long-term average of 0.30.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.3\\\\H_a:\pi\neq 0.3$

The significance level is 0.05.

The sample has a size n=300.

The sample proportion is p=0.399.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.3*0.7}{300}}\\\\\\ \sigma_p=\sqrt{0.0007}=0.026$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.399-0.3-0.5/300}{0.026}=\dfrac{0.097}{0.026}=3.679$

This test is a two-tailed test, so the P-value for this test is calculated as:

$P-value=2\cdot P(z>3.679)=0$

As the P-value (0) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that bearish sentiment is above its long-term average of 0.30.

3 0

3 years ago