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yanalaym [24]
3 years ago
15

Peter planted 5/12 of his garden with asparagus and 3/12 with beans.

Mathematics
1 answer:
UkoKoshka [18]3 years ago
7 0
Answer: He has planted 2/3 and there is 1/3 left to plant.

Explanation: You need to add your fractions together, because each of those is a section of the garden and you need the total of how much of the garden he has planted.

This isn’t too difficult because the denominators are the same.

5/12 + 3/12 = 8/12

It is 8/12 because since the denominators are the same, you just need to add the numerators. Imagine you have a pie that’s cut into 12 pieces, and you and your friends take 5, and then your family takes 3. How many or gone now? 8 pieces. From how many pieces? 12 pieces. So 8/12 pieces are gone.

So Peter has planted 8/12 of his garden. This however, can be simplified, because both of those numbers divide by 4.

8/4 = 2
12/4 = 3

So 8 is now 2, and 12 is now 3.

This is now 2/3.

If there is 2/3 gone, you need to figure out how much is left to get you to 1.

In this instance, 1 can be rewritten as 3/3, because 3 divided by 3 is 1.

In order to get from 2/3 to 1, you need to add 1/3, one more third to the two thirds you already have.

This means Peter has 1/3 left to plant.

Hope this helps :)

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A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life
alisha [4.7K]

Answer:

a)

The null hypothesis is H_0: \mu \leq 10

The alternative hypothesis is H_1: \mu > 10

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Step-by-step explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

H_0: \mu \leq 10

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

H_1: \mu > 10

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

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

10 is tested at the null hypothesis:

This means that \mu = 10

In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.

This means that n = 45, X = 10.5, s = 1.8

Then

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}

t = 1.86

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.

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2 years ago
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