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GREYUIT [131]
3 years ago
10

Marsha recorded the time it took seven children of different ages to run one lap around the track.

Mathematics
2 answers:
max2010maxim [7]3 years ago
7 0
The answer is C. 205.
vlada-n [284]3 years ago
5 0
Answer: C

Reasoning: There is a 40s difference between the age of 4 and 8 meaning 6 most likely would be in the center of that, which also means it’d be about half of 40. So with that we subtract 20 from 225 and thus giving us the closest length of time it should take Marsha’s 6 year old niece to run one lap, 205s.
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Tpy6a [65]

\bf \begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} 40&32\\ 28&16\\ 16&12\\ \cline{1-2} \end{array}~\hfill \stackrel{\textit{average rate of change}}{slope} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{40}~,~\stackrel{y_1}{32})\qquad (\stackrel{x_2}{28}~,~\stackrel{y_2}{16}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{16-32}{28-40}\implies \cfrac{-16}{-12}\implies \boxed{\cfrac{4}{3}} \\\\[-0.35em] ~\dotfill

\bf (\stackrel{x_1}{28}~,~\stackrel{y_1}{16})\qquad (\stackrel{x_2}{16}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-16}{16-28}\implies \cfrac{-4}{-12}\implies \boxed{\cfrac{1}{3}}

8 0
3 years ago
The soccer team and the lacrosse team sold tubs of cookie dough as a fundraiser. Each tub sold earns $5 in profit. If the soccer
Anna [14]
I think it would be 99 i hope this help you
6 0
3 years ago
Read 2 more answers
A. Identify the percentage, rate and base.
Finger [1]

Answer:

1. rate = 30, base = 40, percentage = 12

2. rate = 6, base = 24, percentage = 25

3. rate = 16, base = 64, percentage = 25

4. rate = 4, base = 50, percentage = 20

5. rate = 75, base = 80, percentage = 60

Step-by-step explanation:

5 0
3 years ago
Gina puts $ 4500 into an account earning 7.5% interest compounded continuously. How long will it take for the amount in the acco
Elza [17]

~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5150\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years \end{cases}

5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}

4 0
1 year ago
A restaurant menu has five kinds of soups, seven kinds of main courses, six kinds of desserts, and five kinds of drinks. If a cu
ryzh [129]

Answer: 1050

Step-by-step explanation:

Number of combinations of selecting r things out of n = ^nC_r=\dfrac{n!}{r!(n-r)!}

such that ^nC_1=n

Given: A restaurant menu has 5 kinds of soups, 7kinds of main courses, 6 kinds of desserts, and 5 kinds of drinks.

If a customer randomly selects one item from each of these four categories, then by fundamental counting principle , the number of different outcomes are possible = ^5C_1\times \ ^7C_1\times\ ^6C_1\times\ ^5C_1 =5\times7\times6\times5=1050

hence, total number of outcomes = 1050

3 0
3 years ago
Read 2 more answers
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