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UkoKoshka [18]
3 years ago
8

Evelyn earned a score of 86 on Exam A that had a mean of 71 and a standard deviation of 20. She is about to take Exam B that has

a mean of 550 and a standard deviation of 40. How well must Evelyn score on Exam B in order to do equivalently well as she did on Exam A? Assume that scores on each exam are normally distributed.
Mathematics
1 answer:
Leno4ka [110]3 years ago
4 0

Answer:

580

Step-by-step explanation:

Assuming that the answer should be in terms of z scores, we can calculate the z score as

z = (observed value - mean)/(standard deviation)

For the first exam, the observed value is 86, the mean is 71, and the standard deviation is 20. The z score fot that exam is

z = (86-71)/20 = 0.75

Then, for the second exam, Evelyn has to do equivalently well, so the z score must be the same. Therefore, we have

0.75 = (observed score - 550)/40

multiply both sides by 40 to remove a denominator

0.75 * 40 = observed score - 550

add 550 to both sides to isolate the observed score

0.75 * 40 + 550 = observed score = 580

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sladkih [1.3K]

D is halfway between A and B

so the coordinates of D are (2,2)

E is halfway between A and C so the coordinates of E are (-1,1)

now you need to find the gradient/slope of DE and BC using the formula:

\frac{y2 - y1}{x2 - x1}

<h3><u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>D</u><u>E</u><u>:</u><u> </u></h3>

SUB IN COORDINATES OF D AND E

\frac{1 - 2}{ - 1 - 2}

therefore the gradient of DE is 1/3.

<h3><u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>B</u><u>C</u><u>:</u></h3>

<em>S</em><em>U</em><em>B</em><em> </em><em>I</em><em>N</em><em> </em><em>C</em><em>O</em><em>O</em><em>R</em><em>D</em><em>I</em><em>N</em><em>A</em><em>T</em><em>E</em><em>S</em><em> </em><em>O</em><em>F</em><em> </em><em>B</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>C</em>

<em>\frac{ - 2 - 0}{ - 3 - 3}</em>

therefore the gradient of BC is -2/-6 which simplifies to 1/3.

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6 0
3 years ago
Subtract 7x-9 from2x^2-11
strojnjashka [21]

Answer:

The answer is: 2x^2-7x-2

Step-by-step explanation:

we are asked to subtract 7x-9 from 2x^2-11

so, 2x^2-11-(7x-9)\\=2x^2-11-7x+9\\=2x^2-7x-2.

Hence the desired result is:  2x^2-7x-2.


8 0
3 years ago
Read 2 more answers
the area of a triangle is 124 square units. what would it's new area be if its base was half as long and its height was three ti
Montano1993 [528]
To solve this problem you must apply the proccedure shown below:

 1. You have that the formula for calculate the area of a triangle is:

 A=bh/2

 Where A is the area of the triangle, b is the base of the triangle and h is the height of the triangle.

 bh/2=124
 bh=124x2
 bh=248

 2. The problem asks for the new area of the triangle <span>if its base was half as long and its height was three times as long. Then, you have:

 Base=b/2
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 3. Therefore, when you substitute this into the formula for calculate the area of a triangle, you obtain:

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 4. When you substitute bh=248 into </span>A'=3bh/4, you obtain:
<span>
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<span>
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3 years ago
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High grade steel consists of 85% iron and 15% magnese. Low grade steel consists of 67% iron and 33% mag ese. NASA orders 500 ton
Mamont248 [21]

Answer:

361\dfrac{1}{9} tons of high steel and 138\dfrac{8}{9} tons of low steel

Step-by-step explanation:

Let x be the number of tons of high grade steel and y be the number of tons ow low grade steel needed.

In x tons of high grade steel there are

0.85x tons of iron

0.15x tons of magnese

In y tons of low grade steel there are

0.67y tons of iron

0.33y tons of magnese

NASA orders 500 tons of steel, so

x+y=500

and specifies that it must be in the proportion 80% iron and 20% magnese, so

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From the first equation,

x=500-y

Substitute it into the second equation:

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7 0
3 years ago
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