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Aneli [31]
3 years ago
6

The area of the figure. Use 3.14 for pie

Mathematics
1 answer:
balu736 [363]3 years ago
5 0

Answer:

54.5 ft^{2}

Step-by-step explanation:

divide it to one rectangle and two different triangles.

Rectangles area: (8+3)*4 = 44ft^{2}

triangle top: 4*3/2 = 6ft^{2}

triangle bottom: 3*3/2= 4.5ft^{2}

sum the all up: 44+6+4.5 = 54.5ft^{2}

don't know how to do it with π though if that's obligatory

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2.5833333333333333333

Step-by-step explanation:

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Which ordered pair is the solution
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The answer is D, the ordered pair (5,3) 
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3 years ago
Lets see who is smart lol
Simora [160]

9514 1404 393

Answer:

  \square\quad{y=\dfrac{1}{2}x+4}\\\\\square\quad{\dfrac{y-5}{x-2}=\dfrac{1}{2}}

Step-by-step explanation:

The slope of a line is the same everywhere, so an equation can be written making use of that fact.

  \dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}\\\\\dfrac{y-5}{x-2}=\dfrac{7-5}{6-2}=\dfrac{2}{4}\\\\\boxed{\dfrac{y-5}{x-2}=\dfrac{1}{2}}

Cross-multiplying gives ...

  2(y -5) = x -2

  2y -10 = x -2  . . . eliminate parentheses

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  \boxed{y=\dfrac{1}{2}x+4}

__

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6 0
3 years ago
A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye
Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{x}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

 \sigma_{x}=\sqrt{n}\sigma

The information provided is:

<em>μ</em> = $970

<em>σ</em> = $129

<em>n</em> = 102

Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

P (\sum X \leq  100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}

                              =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

6 0
3 years ago
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