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MAXImum [283]
3 years ago
15

Please answer correctly !!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Olin [163]3 years ago
4 0

U is a point in the line segment TV

TU + UV = TV

TU = 12

UV = 3

<h3>TV = 12+3 = 15 </h3>

<h2><u>TV = 15</u></h2>

Hope this will help...

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5 0
3 years ago
What the percent of 126 is 12
ahrayia [7]
12 is 9.52380952381% of 126
5 0
3 years ago
A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medica
expeople1 [14]

Answer:

The sample mean is of 1925 calories.

The margin of error is of 75 calories.

The sample standard deviation is of 109.7992 calories.

Step-by-step explanation:

Sample mean:

The sample mean is the mean value of the two bounds of the confidence interval. So

M = \frac{1850 + 2000}{2} = 1925

The sample mean is of 1925 calories.

The margin of error

Difference between the bounds and the sample mean. So

2000 - 1925 = 1925 - 1850 = 75 calories.

The margin of error is of 75 calories.

Sample standard deviation:

Here, I am going to expand on the t-distribution.

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 = 18 - 1 = 17

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.898

The margin of error is:

M = T\frac{s}{\sqrt{n}}

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

Since M = 75, T = 2.898, n = 18

M = T\frac{s}{\sqrt{n}}

75 = 2.898\frac{s}{\sqrt{18}}

s = \frac{75\sqrt{18}}{2.898}

s = 109.7992

The sample standard deviation is of 109.7992 calories.

6 0
3 years ago
When would you want to use the median over the mean for describing the measure of center for a data set?
zheka24 [161]
Hello!

Sometimes when there are outliers in a data set, for example 1, 20,25,26,30, one is the outlier and when you find the mean it will not represent the data in the most accurate way. This is when you use the median to represent the part of the data set that has the information that is close together.

I hope this helps!
8 0
3 years ago
Plz plz plz help me with this
erica [24]
This question wants you to find a common denominator for the fractions.
This means finding the LCM, least common multiple, for 21 and 9.
This can be done by listing the multiples for each number until they meet at a common one.

9:
9
18
27
36
45
54
63

21:
21
42
63

This means the LCM of 21 and 9 is 63.
So the lowest possible common denominator is 63.

21 • 3 = 63
So you have to multiply the numerator of 2/21 by 3 as well.
2 • 3 = 6
2/21 = 6/63

Now do the same for 1/9.
9 • 7 = 63
Multiply the numerator, 1, by 7.
1 • 7 = 7
1/9 = 7/63

So in the first blanks, you would put 6/63 for what 2/21 is equal to and 7/63 for what 1/9 is equal to.

7/63 is greater than 6/63.
7/63 > 6/63
That means 1/9 > 2/21
So 2/21 < 1/9 is the answer to the last blank.

Hope this helps!
6 0
3 years ago
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