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arlik [135]
4 years ago
11

Human intelligence is often measured with an Intelligence Quotient (IQ) test. IQ test scores are normally distributed with a mea

n of 100 and a standard deviation of 15. If one person were selected at random to take an IQ test, what is the probability that he or she would score between 90 and 110?
Mathematics
1 answer:
rewona [7]4 years ago
4 0

Answer:

P(90

P(-0.67

P(-0.67

Step-by-step explanation:

Let X the random variable that represent the IQ scores, 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 want to find this probability

P(90

And we can use the z score formula given by:

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

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

P(90

And we can find this probability with this difference:

P(-0.67

And in order to find these probabilities we can use the normal standard table or excel and we got.  

P(-0.67

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swat32
<h3>Answer:  19 dimes</h3>

=================================================

Work Shown:

d = number of dimes

q = 32-d = number of quarters

$5.15 = 515 cents

10d+25q = 515

10d+25(32-d) = 515

10d+800-25d = 515

-15d+800 = 515

-15d = 515-800

-15d = -285

d = -285/(-15)

d = 19

There are 19 dimes. You can stop here if you want.

q = 32-d = 32-19 = 13

There are 13 quarters

-------------------------------

Check:

1 dime = 10 cents

19 dimes = 19*10 = 190 cents

1 quarter = 25 cents

13 quarters = 13*25 = 325 cents

total value = 190 cents + 325 cents = 515 cents = $5.15

The answer is confirmed.

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Can anybody help me???​
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#10

\\ \rm\longmapsto 0

#11

  • Last three i.e Option B,C,D

#13

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\\ \rm\longmapsto y=-|x|+2

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