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choli [55]
3 years ago
5

The distance between Joe’s house and school is 9.25 miles, and Joe goes to school 4 days a week. Estimate the total number of mi

les Joe drives between home and school in one month.
Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
3 0

Answer:

Joe drives about 148 miles each month between  his house and school

Step-by-step explanation:

First of all, we can start by calculating the total number of days in a week that Joe goes to school.

We know that in a standard month, we have about 4 weeks.

Joe goes to school 4 days a week.

Therefore, in a month Joe goes to school for 4 X 4 days = 16 days

The next thing to do is to calculate the total distance travelled by Joe is he travels 9.25 miles each day between his house and the school.

This will be 16 X 9.25 = 148 miles

Joe drives about 148 miles each month between  his house and school

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There are 13 days until her concert.

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The television show CSI: Shoboygan has been successful for many years. That show recently had a share of 18, meaning that among
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Answer:

(a) The value of P (None) is 0.062.

(b) The value of P(at least one) is 0.938.

(c) The value of P(at most one) is 0.253.

(d) The event is not unusual.

Step-by-step explanation:

Let <em>X</em> = number of households watching the show.

The probability of the random variable <em>x</em> is, P (X) = <em>p</em> = 0.18.

The sample selected for the survey is of size, <em>n</em> = 14

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 14 and <em>p</em> = 0.18.

The probability of a Binomial distribution is computed using the formula:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,...

(a)

Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:

P(X=0)={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}=1\times1\times0.06214=0.062

Thus, the value of P (None) is 0.062.

(b)

Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-0.062\\=0.938

Thus, the value of P(at least one) is 0.938.

(c)

Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

             ={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}+{14\choose 1}(0.18)^{1}(1-0.18)^{14-1}\\=0.062+0.191\\=0.253

Thus, the value of P(at most one) is 0.253.

(d)

An event that has a very low probability of occurrence is known as an unusual event.

The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.

This probability value is not low.

Hence, the event is not unusual.

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