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3 years ago

First question- Mean of: 95, 18, 51, 1, 22, 5

2 answers:

8 0

To find the mean, add up the numbers and divide the answer by the number of numbers you added up.

1) 95 + 18 + 51 + 1 + 22 + 5 = 192
192 / 6 = 32
The mean of the numbers is 32.

2) 67 + 103 + 94 + 65 + 18 + 114 + 94 + 63 + 94 + 27 = 739
739 / 10 = 73.9
The mean of the numbers is 73.9.

To find the median, line up all the numbers from least to greatest. Then see which number is in the middle. If there are two numbers in the middle, find the mean of the two numbers.

18, 27, 63, 65, 67, 94, 94, 94, 103, 114
67 and 94 are in the middle. The mean of 67 and 94 is 80.5.
The median is 80.5.

kati45 [8]3 years ago

7 0

The mean is 32
The median is 67

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Write a question that matches this equation.<br><br> 42 ÷ n = 6

Arlecino [84]

If we are supposed to give you like a life situation that matches the equation then I have one:
You have 42 watermelons and you need to divide them into equal groups. Let n equal the number of watermelons in each group. There are 6 groups in total. How many watermelons are in each group? (solve for n).

If this is not what you were looking for then my apologies.

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4 years ago

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

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