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xxMikexx [17]
1 year ago
14

An elevator can hold a maximum of 1500 pounds. An average child weighs 75 pounds and the average adult weighs 150 pounds. The el

evator also can’t fit more than 14 people. How many children and adults can fit in the elevator and stay under the weight limit. Write the inequalities that represent the word problem, and then graph it on DESMOS. What is a solution that represents the amount of children and adults and their combined weight?
Mathematics
1 answer:
CaHeK987 [17]1 year ago
7 0

<u>Let's take this problem step-by-step</u>:

<u>Let's first set up some variables</u>:

  • c: # of children
  • a: # of adults

<u>Let's examine the information given:</u>

  • Elevator can hold a maximum of 1500 pounds

          ⇒ average child is 75 pounds

          ⇒ average adult is 150 pounds

                 ⇒<em> therefore</em>: 75c + 150a\leq 1500

  • Elevator can fit no more than 14 people

          ⇒ <em>therefore</em>: c + a \leq 14

<u>Let's graph the equations</u>:

 75c+150a\leq 1500\\c+a\leq 14

    ⇒ look at the image attached

<u>The point at which the two graphs intersect:</u>

  ⇒ <em>is the solution that represents the amount of children and adults and </em>

<em>       their combine weight</em>

<em />

<u><em>With the horizontal axis being the # of children and vertical axis being the # of adults</em></u><em>:</em>

<em>  ⇒ the </em><em>solution is 8 children and 6 adults</em>

<em></em>

<u>Answer: 8 children and 6 adults</u>

<u></u>

Hope that helped!

<em />

               

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Answer: Choice A.)   88.2 < μ < 93.0

=============================================================

Explanation:

We have this given info:

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  • xbar = 90.6 = sample mean
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  • C = 99% = confidence level

Because n > 30 and because we know sigma, this allows us to use the Z distribution (aka standard normal distribution).

At 99% confidence, the z critical value is roughly z = 2.576; use a reference sheet, table, or calculator to determine this.

The lower bound of the confidence interval (L) is roughly

L = xbar - z*sigma/sqrt(n)

L = 90.6 - 2.576*8.9/sqrt(92)

L = 88.209757568781

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The upper bound (U) of this confidence interval is roughly

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U = 90.6 + 2.576*8.9/sqrt(92)

U = 92.990242431219

U = 93.0

Therefore, the confidence interval in the format (L, U) is approximately (88.2, 93.0)

When converted to L < μ < U format, then we get approximately 88.2 < μ < 93.0 which shows that the final answer is choice A.

We're 99% confident that the population mean mu is somewhere between 88.2 and 93.0

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