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fiasKO [112]
3 years ago
13

Ely’s team scored 62, 76, 53, and 70 points in their first four basketball games. How many points does the team need to score in

the fifth game to have an average of 70 points?
Mathematics
1 answer:
Yanka [14]3 years ago
7 0
89 is the amount of points the team needs to score in the fifth game to average 70 points.

I hope this help
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Answer:

The answer is B.

Step-by-step explanation:

Hope it helps!!

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What is the surface area of the right cone below?
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B) 72pie units squared
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Plz help ASAP which choice is it and why ?
Delicious77 [7]

Answer:

i think it would be choice D

Step-by-step explanation:

this would be because the 20 is 19.8 tons rounded up, the (1.1) is the increase of the original amount plus the 9.8 rounded up, and the (1.05) is the increase by 5.1. the reason why they multiply these like 20(1.1)(1.05) is because each number in the parenthesis is the increase, when you multiply something by 1.1 it will give you it's 100% + the 10% growth, same goes for the 1.05 which will give you the 100% + the 5% of 20(1.1)

in other words if you just multiplied by .1 you would only get 10% of what you're seeking while if you multiply by 1.1 you would get the 110%. the 1 is 100%, and the extra percentage you seek is the decimal (number behind the dot) that follows

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3 years ago
What is 10 as a percent?
Pavel [41]
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7 0
3 years ago
Read 2 more answers
A machine, when working properly, produces 5% or less defective items. Whenever the machine produces significantly greater than
vovikov84 [41]

Answer:

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

Step-by-step explanation:

A machine, when working properly, produces 5% or less defective items.

This means that the null hypothesis is:

H_0: p \leq 0.05

Test if the percentage of defective items produced by this machine is greater than 5%.

This means that the alternate hypothesis is:

H_a: p > 0.05

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.

0.05 is tested at the null hypothesis:

This means that \mu = 0.05, \sigma = \sqrt{0.05*0.95}

A random sample of 300 items taken from the production line contained 27 defective items.

This means that n = 300, X = \frac{27}{300} = 0.09

Value of the test statistic:

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

z = \frac{0.09 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{300}}}

z = 3.18

Pvalue of the test:

Testing if the mean is greater than a value, which means that the pvalue of the test is 1 subtracted by the pvalue of Z = 3.18, which is the probability of a finding a sample proportion of 0.09 or higher.

Looking at the Z-table, Z = 3.18 has a pvalue of 0.9993

1 - 0.9993 = 0.0007

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

5 0
3 years ago
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