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kumpel [21]
3 years ago
8

Yolanda wanted to see if there was a connection between red hair and green eyes. She observed people walking past her on the str

eet and noted their hair and eye color.
Red Hair
Green eyes: 18
Eye color other than green: 29

Hair Color Other Than Red
Green eyes: 114
Eye color other than green: 650

Consider the relative frequency table.

To the nearest whole percent, what is the value of x in the table?

A) x = 14%

B) x = 15%

C) x = 16%

D) x = 18%

Mathematics
2 answers:
spayn [35]3 years ago
7 0

Answer:

Correct choice is A

Step-by-step explanation:

You are given the table

\begin{array}{cccc}&\text{Green eyes}&\text{Other eye color}&\text{Total}\\\text{Red hair}&18&29&47\\\text{Other hair color}&114&650&764\\\text{Total}&132&679&811\end{array}

Find the probabilities:

\begin{array}{cccc}&\text{Green eyes}&\text{Other eye color}&\text{Total}\\\text{Red hair}&\dfrac{18}{811}&\dfrac{29}{811}&\dfrac{47}{811}\\ \\\text{Other hair color}&\dfrac{114}{811}&\dfrac{650}{811}&\dfrac{764}{811}\\ \\\text{Total}&\dfrac{132}{811}&\dfrac{679}{811}&\dfrac{811}{811}\end{array}

Instead of x should be

\dfrac{114}{811}\approx 0.14=14\%.

leva [86]3 years ago
4 0
The answer is x=14% I just took the quiz so.
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If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by y = 70t-16t^2.

To find : The average velocity for the time period beginning when t = 2 and lasting.  a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

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a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}

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b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}

(\text{Average velocity})_{0.01\ s}=7536\ ft/s

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}  

(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}

(\text{Average velocity})_{0.001\ s}=75936\ ft/s

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