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pentagon [3]
3 years ago
5

I will vote u brainliest if u answer so pls do

Mathematics
2 answers:
Lera25 [3.4K]3 years ago
8 0

Answer: For this, you have to calculate the Unit value,

Unit value of 1rst one,  4.99/6 = 0.83

Unit value of 2nd one, 7.99/10 = 0.79

As, 2nd one is cheaper, so it is better :))

Step-by-step explanation:

liberstina [14]3 years ago
8 0

Answer:

Answers are $7.99 for 10 cans and $15.10 for 3 pounds

Step-by-step explanation:

$7.99 for 10 cans is cheaper by .03 and $15.10 for 3 pounds is cheaper by around .25

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Please solve the question below 1
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Answer:

A

Step-by-step explanation:

Find the mean:

Day 1: 2

Day 2: 1

Day 3: 4

Day 4: 5

Day 5: 0

Day 6: 7

Mean = \frac{2 + 1 + 4 + 5 + 0 + 7}{6} = 3.17

This does not exceed a daily mean of 3.5 hours which was set by her mom as a guideline.

Therefore we can conclude that:

"Chawrdi complied with the guideline because the mean number of hours played is \frac{2 + 1 + 4 + 5 + 0 + 7}{6} = 3.17.

8 0
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The Ohio Department of Agriculture tested 203 fuel samples across the state
Rus_ich [418]

Answer:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The proportion estimated would be:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

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PLEASE HELP ASAP! Will give brainliest if you explain
Romashka-Z-Leto [24]
Well, the best way to solve this would involve the Pythagorean theorem. 
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