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Leto [7]
2 years ago
13

STATISTICS in a recent year the mean score on the mathematics section of the SAT test was 515 and the standard deviation was 114

. This means that people within one deviation of the mean have SAT math scores that are no more than 114 points higher or 114 points lower than the mean A. write an absolute value inequality to find the range of SAT mathematics test scores within one standard deviation of the mean B. what is the range of SAT mathematics test scores 12 standard deviation from the mean?
Mathematics
1 answer:
vesna_86 [32]2 years ago
3 0

An absolute value inequality to find the range of SAT mathematics test scores within one standard deviation of the mean is; |x – 515| ≤ 114

<h3>How to Write Inequalities?</h3>

A) We are told that;

Mean score = 515

Standard deviation = 114

We are now given that people within one deviation of the mean have SAT math scores that are no more than 114 points higher or 114 points lower than the mean. Thus, the absolute value inequality is;

|x – 515| ≤ 114

B) The range of scores to within ±2 standard deviations of the mean is;

Range = 515 ± 2(114)

Range = 287 to 743

Read more about Inequalities at; brainly.com/question/25275758

#SPJ1

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56 = −8d + 8 there's not multiple answers
ollegr [7]

Answer:

d= -6

Step-by-step explanation:

56 = -8d + 8

1) You want to get d by itself. Subtract 8 from both sides.

56 = -8d + 8

- 8           - 8

<em>(8 - 8 = 0, and 56 - 8 is 48.)</em>

48 = -8d

2) Then, you divide my -8 on both sides.

(<em>48 / -8 is -6, and -8d / -8 is just d.)</em>

-6 = d

So, d = -6

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4 years ago
I NEED THIS ASAP, PLEASE HELP!!!
WINSTONCH [101]

Answer: 22.5 ; 5 ; 14

Step-by-step explanation:

Given the dataset:

{20,22,23,24,26,26,28,29,30}

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 9

Q1 = 1/4 (9 + 1)th term

Q1 = 1/4(10) = 2.5

We average the 2nd and 3rd term:

(22 + 23) / 2

45 / 2 = 22.5

B) The interquartile range(IQR) of the dataset :

{62,63,64,65,67,68,68,68,69,74}

IQR = Q3 - Q1

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 10

Q1 = 1/4 (10 + 1)th term

Q1 = 1/4(11) = 2.75 term

We take the average of the 2nd and 3rd term:

(63 + 64) / 2

45 / 2 = 63.5

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 10

Q3 = 3/4 (10 + 1)th term

Q3 = 3/4(11) = 8.25 term

We take the average of the 8th and 9th term:

(68 + 69) / 2

137 / 2 = 68.5

IQR = Q3 - Q1

IQR = 68.5 - 63.5

IQR = 5

C) give the dataset :

{7,8,8,9,10,12,13,15,16}

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 9

Q3 = 3/4 (9 + 1)th term

Q3 = 3/4(10) = 7.5 term

We take the average of the 7th and 8th term:

(13 + 15) / 2

28 / 2 = 14

7 0
4 years ago
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Tju [1.3M]

Answer:

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Step-by-step explanation:

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Vinil7 [7]

Answer:

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Step-by-step explanation:

11.

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3 years ago
-4r-8(r-6)=7(5r-1)+8
KATRIN_1 [288]

Answer:

r = 1

Step-by-step explanation:

In this question, we have to solve for the vallue of "r". We will use the distributive property shown below to ease our process.

Distributive Property:

a(b+c) = ab + ac

Now, lets solve this:

-4r-8(r-6)=7(5r-1)+8\\-4r-8r+48=35r-7+8\\-12r+48=35r+1\\48-1=35r+12r\\47=47r\\r=\frac{47}{47}\\r=1

As we have seen the value of r is "1"

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