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Nady [450]
3 years ago
13

Given this expression:

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
3 0

Answer:

2 decimal places

Step-by-step explanation:

just simple count the number of decimal points on both sides

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Q6. (15 points) IQ examination scores of 500 members of a club are normally distributed with mean of 165 and SD of 15.
melamori03 [73]

Answer:

a) 0.8413

b) 421

c) P_{95} = 189.675

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 165

Standard Deviation, σ = 15

We are given that the distribution of  IQ examination scores is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(IQ scores at most 180)

P(x < 180)

P( x < 180) = P( z < \displaystyle\frac{180 - 165}{15}) = P(z < 1)

Calculation the value from standard normal z table, we have,  

P(x < 180) = 0.8413 = 84.13\%

b) Number of the members of the club have IQ scores at most 180

n = 500

\text{Members} = n\times \text{P(IQ scores at most 180)}\\= 500\times 0.8413\\=420.65 \approc 421

c) P(X< x) = 0.95

We have to find the value of x such that the probability is 0.95

P( X < x) = P( z < \displaystyle\frac{x - 165}{15})=0.95  

Calculation the value from standard normal z table, we have,  

P(z < 1.645) = 0.95

\displaystyle\frac{x - 165}{15} = 1.645\\\\x = 189.675  

P_{95} = 189.675

8 0
3 years ago
Solve for N. Express the solution in decimal form.
Dmitriy789 [7]

Answer:

N= 3.8

Step-by-step explanation:

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3 years ago
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Vanyuwa [196]
36:39 in simplest form is 12:13
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The missing credit score was?
Shalnov [3]

Answer:

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Step-by-step explanation:

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College alcohol and drug use: In a Washington Post/Kaiser Family Foundation poll conducted from January through March 2015, 56%
jolli1 [7]

Answer:

d. We are 95% confident that the proportion of adults (ages 17-26) who attended college is between 53% and 59%.

Step-by-step explanation:

Confidence interval of a proportion p.

We build a confidence interval at the x% level from a sample.

With this, we can say that we are x% sure that the true population propotion is in the interval.

The 95% confidence interval is (0.53, 0.59).

This means that we are 95% sure that the true population proportion is between 0.53 and 0.59.

So the correct answer is:

d. We are 95% confident that the proportion of adults (ages 17-26) who attended college is between 53% and 59%.

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