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3 years ago

Dara drove from home to the airport to pick up her

Mathematics

1 answer:

Semmy [17]3 years ago

4 0

Answer:

33 miles

Step-by-step explanation:

<em>Trip to airport:</em>

distance = d1

speed = s1 = 55 mph

time = t1

distance = d2

speed = s2 = 45 mph

time = t2

We are told the same route was used for both trips, so

d1 = d2

The total time was 1 hr 20 min = 1 hr + 20/60 hr = 4/3 hr.

t1 + t2 = 4/3

t2 = 4/3 - t1

speed = distance/time

distance = speed * time

d1 = s1 * t1

d2 = s2 * t2

d1 = d2, so

s1 * t1 = s2 * t2

t2 = 4/3 - t1; s1 = 55; s2 = 45

55 * t1 = 45(4/3 - t1)

55t1 = 60 - 45t1

100t1 = 60

t1 = 0.6

The time to get the the airport was 0.6 hr. Now we find the distance.

d1 = s1 * t1

d1 = 55 * 0.6

d1 = 33

The distance is 33 miles.

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Now we just take square root on both sides of the interval and we got:

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Step-by-step explanation:

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Part a

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On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
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The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

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And replacing into the formula for the interval we got:

$\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

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Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

Part b

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