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Rzqust [24]
3 years ago
15

Suppose the scores on an exam are normally distributed with a mean μ = 75 points, and standard deviation σ = 8 points. Suppose t

hat the top 4% of the exams will be given an A+. In order to be given an A+, an exam must earn at least what score?
Mathematics
1 answer:
alexandr402 [8]3 years ago
6 0

Answer:

89.008

Step-by-step explanation:

z = (x– mean)/ standard deviation

x = z * standard deviation + mean

Now, z value for the top 4% of the exams (which is the same as getting a score below the 96%) has to be found using a z table.

In this case z = 1.751

x = 1.751 * 8 + 75

x = 89.008

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2 years ago
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18=k+12 HELPPP WHATS K
Nana76 [90]

Answer:

k=6

Step-by-step explanation:

18=k+12

you need to isolate k so you subtract 12 from both side

6=k

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There are 20 marbles in a bag. 1/5 are red and others are blue. What is the ratio of red to blue marbles
Levart [38]

Answer:

4:16 or 1:4

Step-by-step explanation:

1/5 of 20 = 4

there are 4 red marbles

there are 16 blue marbles

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2 years ago
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Are a total of 127 Cars And trucks on a lot. If there. For more than twice the number of trucks, then cars, how many cars and tr
kow [346]

Answer:

41 cars and 86 trucks

Step-by-step explanation:

C+T=127

2C+4=T

C+(2C+4)=127

3C+4=127

3C=123

C=41When C=41,

41+T=127

T=127-41

T=86

3 0
3 years ago
A plane has an airspeed of 134 km divided by h. It is flying on a bearing of 78 degrees while there is a 21 km divided by h wind
ANEK [815]

Answer:

Ground speed = 131.25 km/h

Bearing = 92.68º

Step-by-step explanation:

If we imagine a triangle to depict the question with one side 134 km/h at an angle of 78º to the right of vertical (the plane's speed vector).

At the end, we have a line 21 km/h at an angle of 180 (from north), a vertical down from the end of the plane's vector. This vertex is angle B

This makes a triangle with the interior angle B of 78º.

The 3rd side is the ground speed which we can find using the Cosine rule:

b² = a² + c² - 2•a•c•cos(78)

b² = 134² + 21² - 2•134•21•cos(78)

b² = 17226.873

b = 131.25 km/h

The true course bearing" is the ground track. Thus;

Using the Sine rule to find angle A, we have;

(sin(A))/34= sin(78)/b

Thus,

sin(A) = (34 x sin(78))/131.25

sin(A) = 0.2534

A = sin^(-1)0.2534

A = 14.68º

Ground track = A + 78 = 14.68 + 78 = 92.68º

8 0
3 years ago
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