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lukranit [14]
3 years ago
12

Two artists are mixing up green paint for a mural. The first artist, Shane, mixes 3 parts blue and 2 parts yellow to make a shad

e of green. The second artist, Nora, mixes 5 parts blue and 3 parts yellow to make another shade of green. Shane uses 24 ounces of yellow paint in his mixture. How many ounces of blue paint does he use? Explain your reasoning
Mathematics
1 answer:
Marrrta [24]3 years ago
4 0
We would start by breaking down this problem. We want to know how much paint Shane used, so Nora isn't even relevant, we can just throw her out of the equation. Now we know that 2 parts of yellow paint was 24 ounces. We need to determine how much paint is in 1 part:
24/2 = 12
there is 12 ounces of paint in 1 part of paint.
Now what we need to figure out is how much blue paint was used. We know Shane used 3 parts of blue paint, and we know that 1 part = 12 ounces:
3*12 = 36

Shane used 36 ounces of blue paint

Hope this helps.
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An online health and medicine company claims that about 50% of Internet users would go to an online sight first for information
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Answer:

The p-value of the test is 0.0018, which means that for a level of significance higher than this, there is sufficient evidence to claim that less than 50% of Internet users get their health questions answered online.

Step-by-step explanation:

Test the claim that less than 50% of Internet users get their health questions answered online.

At the null hypothesis, we test that 50% of Internet users get their health questions answered online, that is:

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At the alternate hypothesis, we test that this proportion is less than 50%, that is:

H_a: p < 0.5

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.5 is tested at the null hypothesis:

This means that \mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5

A random sample of 1318 Internet users was asked where they will go for information the next time they need information about health or medicine; 606 said that they would use the Internet.

This means that n = 1318, X = \frac{606}{1318} = 0.4598

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z = \frac{0.4598 - 0.5}{\frac{0.5}{\sqrt{1318}}}

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P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.4598, which is the p-value of z = -2.92.

Looking at the z-table, z = -2.92 has a p-value of 0.0018.

The p-value of the test is 0.0018, which means that for a level of significance higher than this, there is sufficient evidence to claim that less than 50% of Internet users get their health questions answered online.

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