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Usimov [2.4K]
3 years ago
7

Bob is packing a suitcase to take on an airplane. A suitcase and its contents can weigh at most 50 pounds before the airlines ch

arge extra fees. Bob's suitcase weighs 10 pounds when it is empty, so he can put 40 pounds of clothes in the suitcase. Bob's shirts weigh 0.5 pounds each and his pants weigh 2 pounds each.
**
Write an equation that would model how many pounds of clothing Bob can pack.

Rewrite the equation to model the number of pants Bob can pack in terms of the number of shirts.
Mathematics
2 answers:
bagirrra123 [75]3 years ago
7 0
He is correct u can trust him
Blizzard [7]3 years ago
4 0
The answer to the shirt is 0.5x+10=50 and the answer to the pants equation
2x+10=50
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In △ABC, point D∈ AB with AD:DB=5:3, point E∈ BC so that BE:EC=1:4. If AABC=40 in2, find AADC, ABDC, and ACDE.
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8 0
4 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.

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3 years ago
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