2(3) squared answer and steps.

Your parents give you 1 cent to clean up your kitchen on Aug. 15. They will double the amount of money they give you on the foll

kotegsom [21]

Answer:

Step-by-step explanation:

Aug

15 = 1 26 = 23.04 Total $737.28 that's a great deal

16 = 2 27 = 46.08

17 = 4 28 = 92.16

18 = 8 29 = 184.32

19 = 16 30 = 368.64

20 = 36 31 = 737.28

21 = 72

22 = 1.44

23 = 2.88

24 = 5.76

25 = 11.52

7 0

3 years ago

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Geometry help will give brainliest answer the blank ones

stiks02 [169]

3) Corresponding angles
4) Alternate exterior angles
5) Corresponding angles
7) Supplementary angles
9) Same side interior angles
10) Alternate exterior angles

If you need more explanations, let me know

4 0

3 years ago

Determine the solution to the following system of equations:

anyanavicka [17]

Answer:

no solution to the question

5 0

3 years ago

What is 7x =90+22 ?

Elanso [62]

Answer:

x=16

Step-by-step explanation:

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3 years ago

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Membership in Mensa requires an IQ score above 131.5. Nine candidates take an IQ test, and they have a mean IQ score of 133. IQ

rodikova [14]

Answer:

a) $P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

$P(Z\geq 2.2)=1- P(z$

b) $P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

And we can find this probability using the complement rule:

$P(Z\geq 6.6)=1- P(z$

c) No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

$P(Z\leq 6.3) \approx 1$

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

A. If 1 person is randomly selected from thegeneral population, find the probability of getting someone with anIQ score of at least 133.

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(100,15)$

Where $\mu=100$ and $\sigma=15$

We are interested on this probability

$P(X\geq 133)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

$P(Z\geq 2.2)=1- P(z$

And the probability is calculated from the normal standard table or with excel.

B. If 9 people are randomly selected,find the probability that their mean IQ score is at least 133.

For this case since the distribution for the random variable X is normal then the distribution for the sample mean is also normal and given by:

$\bar X = \sim N(\mu= 100 ,\sigma_{\bar x}= \frac{15}{\sqrt{9}}=5)$

The new z score is defined as

$z=\frac{x-\mu}{\sigma_{\bar x}}$

If we apply this formula to our probability we got this:

$P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

And we can find this probability using the complement rule:

$P(Z\geq 6.6)=1- P(z$

And the probability is calculated from the normal standard table or with excel.

C. Although the results are available,the individual IQ test scores have been lost. Can it be concluded that all 9 candidates have IQ scores above 131.5 so that they allare eligible for Mensa membership?

No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

$P(Z\leq 6.3) \approx 1$

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3 years ago

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