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Anika [276]
3 years ago
11

What is 6×8 and 11×11

Mathematics
2 answers:
Vanyuwa [196]3 years ago
8 0
Hi there


6*8 = 48
and
11 * 11 = 121

I hope that's help !
Margaret [11]3 years ago
6 0
First answer is 48
Second answer is 121




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According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones
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Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

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(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

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We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = <u><em>back-to-college family spending on electronics</em></u>

SO, X ~ Normal(\mu=237,\sigma^{2} =54^{2})

The z score probability distribution for normal distribution is given by;

                                 Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean family spending = $237

           \sigma = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

        P(X < $150) = P( \frac{X-\mu}{\sigma} < \frac{150-237}{54} ) = P(Z < -1.61) = 1 - P(Z \leq 1.61)

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        P(X > $390) = P( \frac{X-\mu}{\sigma} > \frac{390-237}{54} ) = P(Z > 2.83) = 1 - P(Z \leq 2.83)

                                                             = 1 - 0.9977 = <u>0.0023</u>

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

     P($120 < X < $175) = P(X < $175) - P(X \leq $120)

     P(X < $175) = P( \frac{X-\mu}{\sigma} < \frac{175-237}{54} ) = P(Z < -1.15) = 1 - P(Z \leq 1.15)

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Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>

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