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vekshin1
1 year ago
7

Iq scores are known to be normally distributed. the mean iq score is 100 and the standard deviation is 15. what percent of the p

opulation has an iq over 115?
Mathematics
1 answer:
kobusy [5.1K]1 year ago
6 0

The 100%-Norm.Dist(value,mean,sd,true) percent of the population has an iq over 115

According to the statement

we have to find that the what percent of the population has an iq over 115.

And from given information,

the mean iq score is 100 and the standard deviation is 15.

And

Normal distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

And by use of this formula, we find the percentage of the population has an iq over 115.

Then the result will comes as 100%.

hence, 100%-Norm.Dist(value,mean,sd,true)

So, The 100%-Norm.Dist(value,mean,sd,true) percent of the population has an iq over 115.

Learn more about Normal distribution here

brainly.com/question/4079902

#SPJ4

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On May 17th Jane took out a loan for $33,000 at 6% to open her law practice office the loan will mature the following year on Ja
frosja888 [35]

Answer:

$ 31050

Step-by-step explanation:

<em>Step 1 : Write the formula for calculating simple interest.</em>

Simple Interest = <u>P x R x T </u>

                                100

P: Principal Amount-The loan taken (30,000)

R: Interest rate at which the loan is give (6)

T: Time period of the loan in years-there are 12 months in 1 year. There are 7 months from May till June (7/12)

<em>Step 2: Substitute values in the formula</em>

Simple Interest = <u>30,000 x 6 x 7/12</u>

                                       100

Simple Interest = $1050

<em>Step 3: Calculate the amount due at maturity</em>

At the maturity or the end of the time period given, the original or principal amount of the loan has to be repaid along with the simple interest.

Amount at maturity = Principal Amount + Simple Interet

Amount at maturity = 30,000 + 1050

Amount at maturity = $31050

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Step-by-step explanation:

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Fiona wrote the linear equation y = 2/5 x – 5. When Henry wrote his equation, they discovered that his equation had all the same
Hoochie [10]

we have

y=\frac{2}{5}x-5

Let's Isolate the variable y in each of the cases and then compare with Fiona's equation  to determine the solution of the problem

case A) x-\frac{5}{4}y= \frac{25}{4}

Multiply by 4 both sides

4x-5y= 25

5y=4x-25

Divide by 5 both sides

y=\frac{4}{5}x-5

therefore

the case A) is not equal to Fiona's equation

case B) x-\frac{5}{2}y= \frac{25}{4}

Multiply by 4 both sides

4x-10y= 25

10y=4x-25

Divide by 10 both sides

y=\frac{2}{5}x-2.5

therefore

the case B) is not equal to Fiona's equation

case C) x-\frac{5}{4}y= \frac{25}{2}

Multiply by 4 both sides

4x-5y= 50

5y=4x-50

Divide by 5 both sides

y=\frac{4}{5}x-10

therefore

the case C) is not equal to Fiona's equation

case D) x-\frac{5}{2}y= \frac{25}{2}

Multiply by 2 both sides

2x-5y= 25

5y=2x-25

Divide by 5 both sides

y=\frac{2}{5}x-5

therefore

the case D) is  equal to Fiona's equation

<u>the answer is</u>

x-\frac{5}{2}y= \frac{25}{2}



7 0
2 years ago
Read 2 more answers
A quality-assurance inspector periodically exam-ines the output of a machine to determine whether it is properly adjusted. When
Stolb23 [73]

Answer:

Step-by-step explanation:

From the given information;

The null hypothesis & alternative hypothesis:

\mathbf{H_o:\mu = 2}  \\ \\  \mathbf{H_1 : \mu \ne 2}

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}

Z = \dfrac{2.025- 2}{\dfrac{0.07}{\sqrt{35}}}

Z = \dfrac{0.025}{\dfrac{0.07}{\sqrt{35}}}

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The p-value = 2(Z > 2.11)  since this is a two-tailed test

The p-value = 2( 1 - Z < 2.11)

The p-value = 2 (1 -0.9826)

The p-value = 2 (0.0174)

The p-value = 0.0348

Decision Rule: To reject the \mathbf{H_o}  if the p-value is less than \mathbf{H_o}

Conclusion: We fail to reject \mathbf{H_o} and conclude that the population mean = 2, thus the machine is properly adjusted.

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