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Masja [62]
3 years ago
13

What is the solution to the inequality? −4(x−6)≤−2x+6

Mathematics
1 answer:
Kipish [7]3 years ago
3 0
-4(x-6) ≤ - 2x+6
-4x-24≤ -2x+6
+2x +2x
-2x-24≤+6
+24 +24
-2x ≤ 30
Divide both sides by -2
and you get....
x ≥ -15
You switch the inequality sign because you're dividing by a negative.
I hope all is well, and you pass! Good luck, rockstar! (:
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The Math Club is raising money so they can attend a local competition next month. They have already raised $125 and they are sel
prisoha [69]

Answer: 100 candy bars

Step-by-step explanation:

You did not include the question but judging by the progression I can assume that the question is along the lines of how many candy bars they will need to sell to reach their target.

You did not include the amount they have to raise as well so let us assume that number is $375.

Assuming they have to raise $375 and they have already raised $125, the amount left to be raised is:

= 375 - 125

= $250

With each candy bar selling at $2.50, the number of candy bars they will need to sell is:

= 250/ 2.50

= 100 candy bars

<em>You should use this process to find the answer to your question. </em>

3 0
3 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
Help please asap!!!!!!
Igoryamba
I hope this helps you :)

4 0
3 years ago
The data below show the number of games won by a football team in each of the last 15 seasons. What is a histogram that represen
Irina-Kira [14]
A sample histogram is attached.

Using intervals of width 5, we have
0-4, with a frequency of 6
5-9, with a frequency of 4
10-14, with a frequency of 3
15-19, with a frequency of 2

8 0
3 years ago
K+1/2 =4 solve for the variable​
mina [271]

Answer:

k+1/2 =4

subtract 1/2 from both sides

= k = 3.5 or 3 1/2

4 0
2 years ago
Read 2 more answers
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