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NeX [460]
4 years ago
7

One of the events at the circus starred Gabriella the Human cannonball. On Saturday she performed in four shows. Shot from a can

non, her distances measured 8.05 meters, 17.71, 14.38 meters, and 19 35 meters. What was the total distance Gabriella flew that day
Mathematics
1 answer:
Amanda [17]4 years ago
4 0

Answer:

59.49 m

Step-by-step explanation:

We are given that Gabriella performed in four shows on Saturday in an event of the circus.

We have to find total distance  flew by Gabriella on that day

In first shot from cannon , Gabriella flew distance=8.05 m

In second shot form cannon,Gabriella flew distance=17.71 m

In third shot from cannon, Gabriella flew distance=14.38 m

In fourth shot from cannon,Gabriella flew distance=19.35 m

Total distance flew by Gabriella on that day=8.05+17.71+14.38+19.35=59.49 m

Hence, Gabriella flew total distance on that day =59.49 m

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The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. T
Anit [1.1K]

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a) 25

b) 67

c) 97

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample. In this problem, \sigma = 0.25

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

M = z*\frac{\sigma}{\sqrt{n}}

0.1 = 1.96*\frac{0.25}{\sqrt{n}}

0.1\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.1}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}

n = 24.01

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

M = z*\frac{\sigma}{\sqrt{n}}

0.06 = 1.96*\frac{0.25}{\sqrt{n}}

0.06\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.06}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}

n = 66.7

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

M = z*\frac{\sigma}{\sqrt{n}}

0.05 = 1.96*\frac{0.25}{\sqrt{n}}

0.05\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.05}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}

n = 96.04

Rounding up, 97

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