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Molodets [167]
3 years ago
9

Suppose SAT Writing scores are normally distributed with a mean of 498 and a standard deviation of 108. A university plans to aw

ard scholarships to students whose scores are in the top 4%. What is the minimum score required for the scholarship? Round your answer to the nearest whole number, if necessary.

Mathematics
1 answer:
Katen [24]3 years ago
5 0

Answer: 687

Step-by-step explanation:

normally distributed

μ = 498

σ = 108

Top 4%

P(X > a) = 0.04

P(Z > a-498/108) = 0.04

Using the table attached, for the top 4%:

0.50 - 0.04 = 0.46

a-498/108 = 1.75

a-498 = 189

a = 687

The minimum score for the scholarship is 687.

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Jeremy find the cost by adding the percents. 25% off of $60 is equal to 15% off of $60, then subtracting 10% from that cost is e
bonufazy [111]

Answer:

a. 25% off of $60 is equal to $45.

b. 15% off of $60, then subtracting 10% from that cost is equal to $45.

c. The two total costs are $45 and $45, and their addition is equal to $90 (i.e. $45 + $45 = $90).

Step-by-step explanation:

These can be determined as follows:

a. Calculation of the first total cost, i.e. 25% off of $60

25% of 60 = 25% * $60 = 15

First total cost = 25% off of $60 = $60 - (25% of $60) = $60 - $15 = $45

Therefore, 25% off of $60 is equal to $45.

b. Calculation of the second total cost, i.e. 15% off of $60, then subtracting 10% from that cost

15% of $60 = $9

10% of $60 = $6

Cost = 15% off of $60 = $60 - (15% of $60) = $60 - $9 = $51

Second total cost = Cost - (10% of $60) = $51 - $6 = $45

Therefore, 15% off of $60, then subtracting 10% from that cost is equal to $45.

c. The two total costs

The two total costs are 25% off of $60 which is equal to $45 and 15% off of $60, then subtracting 10% from that cost which is also equal to $45. The addition of the two costs is $90 (i.e. $45 + $45 = $90).

Therefore, the two total costs are $45 and $45, and their addition is equal to $90 (i.e. $45 + $45 = $90).

7 0
3 years ago
Which model represents the expression
NARA [144]

Answer:

A

Step-by-step explanation:

Because you add a denominator of 1 to the whole number and just multiply over and you get 6/8

6 0
2 years ago
Find angle B if triangle ABC below is isosceles.
crimeas [40]

2x-20 = x+8

x-20 =8

x = 28

28+8 =36

 angle B = 36 degrees

answer is b

5 0
3 years ago
5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g
anygoal [31]

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1117}{10} = 111.7

Sum of squares of differences = 642.1

S.D = \sqrt{\frac{642.1}{49}} = 8.44

We are given the following in the question:  

Population mean, μ = 106

Sample mean, \bar{x} = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

H_{0}: \mu = 106\\H_A: \mu > 106

We use one-tailed(right) t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135

Now,

t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833

Since,                  

t_{stat} > t_{critical}

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

4 0
2 years ago
Pls owrk out the missing angle
Alik [6]

Answer:

125

Step-by-step explanation:

A full circle is 360°

80+85+70=235

360-235=125

4 0
2 years ago
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