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Debora [2.8K]
3 years ago
14

Anna drives 15,500 miles a year for work. On average, how many miles does she drive a week?

Mathematics
2 answers:
sergiy2304 [10]3 years ago
6 0
298 miles. there is a total of 52 weeks a year so divide 15,500 by 52 to get miles per week
I am Lyosha [343]3 years ago
3 0

Answer:

206 miles

Step-by-step explanation:

People generally work 5x a week so.. (15000 / 52)

52 indicates the amount of weeks in a year.

In total, that is 288.4 miles a week. Now, we're assuming she doesn't work on the weekend so we divide 288.4 by 7. That is 41.2 miles a day. Now, you times 41.2 by 5 and your answer should be 206 miles a week.

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Read 2 more answers
A random variable X is uniformly distributed over the interval (48, 96).
Katen [24]

Answer:

a) P(50 < X

b) P(X

c) P(X>90)=1- \frac{90-48}{48} = 1-0.875 = 0.125

d) P(X

And using the CDF we got:

P(X

P(X>80)

And using the CDF and the complement rule we got:

P(X>80)=1- \frac{80-48}{48} = 1-0.667 = 0.333

So then the probability required would be:

P(X < 60 or X > 80) = 0.25+0.333= 0.583

Step-by-step explanation:

For this case we define the random variable of interest X, and we know the distribution given by:

X \sim Unif (a= 48, b=96)

The density function is given by:

f(x) = \frac{1}{96-48} = \frac{1}{48} , 48 \leq X \leq 96

And the cumulative distribution function is given by:

F(x) = \frac{x-48}{96-48} , 48 \leq X \leq 96

Part a

We want this probability:

P(50 < X

And we can find this probability with this difference:

P(50 < X

And replacing we got:

P(50 < X

Part b

We want this probability:

P(X

And using the CDF we got:

P(X

Part c

We want this probability:

P(X>90)

And using the CDF and the complement rule we got:

P(X>90)=1- \frac{90-48}{48} = 1-0.875 = 0.125

Part d

P(X

And using the CDF we got:

P(X

P(X>80)

And using the CDF and the complement rule we got:

P(X>80)=1- \frac{80-48}{48} = 1-0.667 = 0.333

So then the probability required would be:

P(X < 60 or X > 80) = 0.25+0.333= 0.583

8 0
3 years ago
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