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2 years ago

I will give brainliest help asap

Mathematics

1 answer:

just olya [345]2 years ago

8 0

it 22................

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If the heights of a population of men are approximately Normally distributed, and the middle 99.7% have heights between 5'0" and

kirza4 [7]

Answer:

4 inches

Step-by-step explanation:

the middle number is 6 and the distance from 99.7 from the middle is 3 standard deviations. 12/3 =4

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2 years ago

1.The sum of two times a number and 8 is 10. What is the number?<br> A.1<br> B.2<br> C.18<br> D.9

Soloha48 [4]

Answer is Choice A - 1.
1 times 2 is 2, and 2 + 8 is 10.

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2 years ago

Mrs.gauthier plans to take her class on 2 field trips this year there are 23 students in her class and each field trip will cost

Alina [70]

If there is 2 field trips with 23 students and each student costs $5 dollars. So for the 1st trip it will be 23x$5

1st trip- $115

And on the second trip its the same because there is 23 students that cost 5 dollars so it will be 23x5

2nd trip- $115

To find out the total of both trips youd add 115+115/

In total, for both trips it will cost $230 dollars.

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3 years ago

The double-bar graph shows the win-loss records for the Carolina Panthers football team in the years 1995-2001. A bar graph titl

lina2011 [118]

Answer:

the answer is A

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Six different second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic re

Yanka [14]

Answer:

$Range=14$

$\sigma^2 =32.4$

$\sigma = 5 .7$

The standard deviation will remain unchanged.

Step-by-step explanation:

Given

$Data: 136, 129, 141, 139, 138, 127$

Solving (a): The range

This is calculated as:

$Range = Highest - Least$

Where:

$Highest = 141; Least = 127$

So:

$Range=141-127$

$Range=14$

Solving (b): The variance

First, we calculate the mean

$\bar x = \frac{1}{n} \sum x$

$\bar x = \frac{1}{6} (136+ 129+ 141+ 139+ 138+ 127)$

$\bar x = \frac{1}{6} *810$

$\bar x = 135$

The variance is calculated as:

$\sigma^2 =\frac{1}{n-1}\sum(x - \bar x)^2$

So, we have:

$\sigma^2 =\frac{1}{6-1}*[(136 - 135)^2 +(129 - 135)^2 +(141 - 135)^2 +(139 - 135)^2 +(138 - 135)^2 +(127 - 135)^2]$

$\sigma^2 =\frac{1}{5}*[162]$

$\sigma^2 =32.4$

Solving (c): The standard deviation

This is calculated as:

$\sigma = \sqrt {\sigma^2 }$

$\sigma = \sqrt {32.4}$

$\sigma = 5 .7$ --- approximately

Solving (d): With the stated condition, the standard deviation will remain unchanged.

5 0

3 years ago