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3 years ago

Use the graph to complete the following. The probability that a boxer weighs between 60 and 80 pounds is %

Mathematics

2 answers:

AlexFokin [52]3 years ago

7 0

Answer:

95%

Step-by-step explanation:

The Empirical rule, also the 68–95–99.7 rule, states that for a population that is approximately normal or symmetrical, nearly all of the data values will lie within three standard deviations of the mean;

68% of data values will fall within one standard deviation from the mean

95% of data values will fall within two standard deviation from the mean

99.7% of data values will fall within three standard deviation from the mean

From the graph given, we note that the weights 60 and 80 pounds fall within two standard deviations from the mean;

70 ± (2*5) = 70 ± 10 = (60, 80)

70 is the mean, 5 the standard deviation and 2 the number of standard deviations from the mean. From the Empirical rule we can conclude that the probability that a boxer weighs between 60 and 80 pounds is 95%

Alona [7]3 years ago

7 0

Answer:

95% on EDGE

Step-by-step explanation:

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Marrrta [24]

Answer:

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Step-by-step explanation:

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2 years ago

a produce store sells 63 apples if the ratio of red apples to green apples was sold was 7 to 2 what is the combined amount of re

Hitman42 [59]

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4 0

3 years ago

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A grocer stacks oranges in a four-sided pyramid that is 7 layers high. How many oranges are in the pile?

gayaneshka [121]

Short Answer D
P(1) = 1(1+1)(2*1 + 1)/6
P(1) = 1(2)(2 +1) / 6
P(1) = 1(2)(3)/6
P(1) = 1

P(2) = 2(2+1)(2*2 + 1) / 6
P(2) = 2(3)(5) / 6
P(2) = 5 So this formula is adding as it goes along. To Find the Total all we need do is use the formula to calculate P(1) to P(7)

P(7) = 7*(7 + 1)(2*7 + 1)/6
P(7) = 7 * 8 * 15 / 6
P(7) = 7 * 4 * 5
P(7) = 140 <<<< Answer

4 0

3 years ago

HELP

Westkost [7]

Answer:

see below

Step-by-step explanation:

The exponent rules that apply are ...

(a^b)(a^c) = a^(b+c)

a^-b = (1/a)^b

(a^b)^c = a^(b·c)

_____

These let you rewrite the given function as ...

f(x) = (3^(2x))(3^1) = 3(3^(2x)) = 3(3^2)^x = 3·9^x

and

f(x) = 3^(2x+1) = (3^-1)^(-(2x+1)) = (1/3)^-(2x+1)

5 0

3 years ago

I'm sorry but these are the multiple choice answers.

Sunny_sXe [5.5K]

$\bf \qquad \qquad \textit{Future Value of an ordinary annuity}
\\\\
A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]$

$\bf \qquad 
\begin{cases}
A=
\begin{array}{llll}
\textit{original amount}\\
\textit{already compounded}
\end{array}&
\begin{array}{llll}

\end{array}\\
pymnt=\textit{periodic payments}\to &200\\
r=rate\to 7\%\to \frac{7}{100}\to &0.07\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, four quarters}
\end{array}\to &4\\

t=years\to &12
\end{cases}
\\\\\\
A=200\left[ \cfrac{\left( 1+\frac{0.07}{4} \right)^{4\cdot 12}-1}{\frac{0.07}{4}} \right]$

7 0

3 years ago