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Keith_Richards [23]
3 years ago
11

Look in the comments to see the question :)

Mathematics
1 answer:
Maru [420]3 years ago
3 0
I’m trying to looking :)
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The sum of two numbers makes thirteen and their difference is 7
Anna35 [415]

Answer:

10 and 3

Step-by-step explanation:

10-3=7

10+3=13

8 0
3 years ago
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It costs $7.00 to purchase a
rusak2 [61]

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12.25 U.S. dollars

Step-by-step explanation:

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An orchard has 418 apple trees. The number of rows exceeds the number of trees per row by 3. How many trees are there in each ro
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There are about 139 trees in each row.
418/3=
139.333333333~ 139
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Suppose a life insurance company sells a ​$250 comma 000 ​one-year term life insurance policy to a 23​-year-old female for ​$190
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Answer:300000

Step-by-step explanation:

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4 0
3 years ago
In a survey, 24 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
erica [24]

Answer:

The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 24 - 1 = 23

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 23 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.98}{2} = 0.99. So we have T = 2.5

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.5\frac{2}{\sqrt{24}} = 1.02

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.02 = $40.98.

The upper end of the interval is the sample mean added to M. So it is 42 + 1.02 = $43.02.

The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.

7 0
3 years ago
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