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Artemon [7]
3 years ago
13

The mean on a standardized test is 100 and the standard deviation is 35. Your score is 65. What percentage of the scores were hi

gher than yours?
Mathematics
1 answer:
Helen [10]3 years ago
6 0

Answer:

84.13%

Step-by-step explanation:

Given data:

mean μ= 100 and standard deviation= 35

P(x>65)=P(z>\frac{\mu-\sigma}{\sigma})

putting vales we get

P(z>(65-100)/35)=P(z>-1) or P(z<1).,

from normal distribution table we get

P(z<1)= 0.8413= 84.13%

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Aria played 2 games and went on 7 rides.

Step-by-step explanation:

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3 years ago
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A universal set U consists of elements. If sets​ A, B, and C are proper subsets of U and ​n(U)​, ​n(A ​B)​n(A ​C)​n(B ​C)​, ​n(A
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Answer:

n(A\ u\ B) = 10

n(A'\ u\ C) = 10

n(A\ n\ B)' = 6

Step-by-step explanation:

Given

n(U) = 12

n(A\ n\ B) =n(A\ n\ C) = n(B\ n\ C) =6

n(A\ n\ B\ n\ C)=4

n(A\ u\ B\ u\ C)=10

Required

Solve a, b and c

There are several ways to solve this; the best is by using Venn diagram (see attachment for diagram)

Solving (a):

n(A\ u\ B)

This is calculated as:

n(A\ u\ B) = n(A) + n(B) - n(A\ n\ B)

From the attachment

n(A) = 0+2+4+2 = 8

n(B) = 0+2+4+2 = 8

n(A\ n\ B) = 4 +2 = 6

So:

n(A\ u\ B) = 8 + 8 - 6

n(A\ u\ B) = 10

Solving (b):

n(A'\ u\ C)

This is calculated as:

n(A'\ u\ C) = n(A') + n(C) -n(A'\ n\ C)

From the attachment

n(A) = n(U) - n(A) = 12 - 8 = 4

n(C) = 0+2+4+2 = 8

n(A'\ n\ C) = 2

So:

n(A'\ u\ C) = 4 + 8 - 2

n(A'\ u\ C) = 10

Solving (c):

n(A\ n\ B)'

This is calculated as:

n(A\ n\ B)' = n(U) - n(A\ n\ B)

n(A\ n\ B)' = 12- 6

n(A\ n\ B)' = 6

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Evaluate 15/r -1 when r = 5
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\frac{15}{r}-1=\frac{15}{5}-1=3-1=2

2)
\frac{15}{r-1}=\frac{15}{5-1}=\frac{15}{4}=3\frac{3}{4}=3.75
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Answer:

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