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sleet_krkn [62]
3 years ago
15

A class’s test scores are normally distributed if the average score 60 and the standard deviation is 7 choose the spots where th

e students scoring below 46 and above 74 lie
Mathematics
2 answers:
AleksandrR [38]3 years ago
6 0

Answer: Look at the picture below

Step-by-step explanation: I got this right on Edmentum

Brrunno [24]3 years ago
5 0

Answer:

46

Step-by-step explanation:

We are given that a  class's test scores are normally distributed with an average score of 60.

We know that the curve of a normal distribution is symmetric about its mean.

60-14=46

60+14=74

Hence, the point 46 lies to the left of the mean and 74 lies to the right of the mean, and the two points have the same function value.

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Bingel [31]
If the perimeter is 16, the side length is 4 units.

To find the area of the shaded sections, you will use the fractional part that each section represents and multiply that fraction by the total area of the square.

Area of DEA is 1/4 of 16 square Units (4 x 4).

1/4 x 16 = 4 square units

Area of EFB is 1/8 of 16 square units.

1/8 x 16 = 2 square units

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The area of the shaded region is 6 square units.

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3 years ago
Pleas help if possible :)
Ronch [10]

Answer: i don’t know but good luck, you got this!Have a nice day!

Step-by-step explanation:

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Fynjy0 [20]

Answer:

x = 41

Step-by-step explanation:

(2x+20) + (2x-4) = 180 degrees

4x+16 = 180 degrees

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If Jon has 26 apples and ate four how many does he have
netineya [11]
22 is it’s the basic
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3 years ago
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Solve the triangle that has a=4.6, B=19°, A=92° (picture provided)
kolezko [41]

Answer:

Option b

Step-by-step explanation:

To solve this problem use the law of the sines.

We have 2 angles of the triangle and one of the sides.

a = 4.6\\B = 19\°\\A = 92\°\\C = 180 -A - B\\C = 180 - 92 - 19\\C = 69\°

The law of the sines is:

\frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c}

Then:

\frac{sin(92)}{4.6} = \frac{sin(19)}{b}\\\\b = \frac{sin(19)}{\frac{sin(92)}{4.6}}\\\\b = 1.5

\frac{sin(B)}{b} = \frac{sin(C)}{c}\\\\\frac{sin(19)}{1.5} = \frac{sin(69)}{c}\\\\c = \frac{sin(69)}{\frac{sin(19)}{1.5}}\\\\c = 4.30

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3 years ago
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