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mina [271]
2 years ago
3

A cargo helicopter delivers only 100-pound packages and 120-pound packages. For each delivery trip, the helicopter must carry at

least 10 packages, and the total weight of the packages can be at most 1,100 pounds. What is the maximum number of 120-pound packages that the helicopter can carry per trip?

Mathematics
2 answers:
vaieri [72.5K]2 years ago
5 0

Answer:

  5

Step-by-step explanation:

A graphical solution shows the number of 120-pound packages can be at most 5.

_____

In the graphical solution, the doubly-shaded area is the "feasible region". Its maximum extent in the "120-pound packages" direction is 5.

Shalnov [3]2 years ago
5 0
<h2>Maximum number of 120 pound packages helicopter can carry is 5.</h2>

Step-by-step explanation:

Let m be the number of 100 pond packages and n be the number of 120 pound packages.

For each delivery trip, the helicopter must carry at least 10 packages.

          m + n ≥ 10

          m ≥ 10 - n

Total weight of the packages can be at most 1,100 pounds

         100m + 120 n ≤ 1100  

         10 m + 12 n  ≤ 110

         5 m + 6 n  ≤ 55  

          5 (10 - n ) + 6 n  ≤ 55

          50 - 5n + 6 n ≤ 55

                      n ≤ 5

So maximum number of 120 pound packages helicopter can carry is 5.

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Please help me answer this ASAP! The two-way table shows the number of ninth and tenth graders who prefer going to sporting even
Elis [28]
<h2>Answer</h2>

0.43

<h2>Explanation</h2>

Remember that RelativeFrecuency=\frac{Frecuency}{SumOfAllFrecuencies}

Since the problem is telling us "Among tenth graders", we must focus on the 10th graders row only. From the row, we can infer that the frequency is the number of 10th graders who prefer going to sporting events, so Frecuency=6. Now, the sum of all frequencies will be the sum of all the 10th graders, so SumOfAllFrecuencies=6+8=14. Let's replace the values:

RelativeFrecuency=\frac{Frecuency}{SumOfAllFrecuencies}

RelativeFrecuency=\frac{6}{14}

RelativeFrecuency=0.4285

And rounded to the nearest hundredth:

RelativeFrecuency=0.43

4 0
2 years ago
A population of 550 rabbits is increasing by 7.5% each year. in about how many years will the population be over 1000?
Furkat [3]

I believe this would take the form of an exponential equation:

A = Ao (1 + r)^t

where A is final population, Ao is initial population, r is rate of growth and t is time

 

A / Ao = (1 + r)^t

log A / Ao = t log (1 + r)

t = (log A / Ao) / log (1 + r)

t = [log (1000 / 550)] / log (1.075)

t = 8.27 years

 

SO the answer is B) about 9 years

5 0
2 years ago
A scatterplot is produced to compare the size of a lake to the number of fish that are in it. There are 15 data points, each rep
topjm [15]
There is no relation between the size of a lake and the number of fish in the lake.
8 0
2 years ago
Read 2 more answers
The Acme Candy Company claims that​ 60% of the jawbreakers it produces weigh more than 0.4 ounces. Suppose that 800 jawbreakers
vichka [17]

Answer:

Yes, it would be statistically significant

Step-by-step explanation:

The information given are;

The percentage of jawbreakers it produces that weigh more than 0.4 ounces = 60%

Number of jawbreakers in the sample, n = 800

The mean proportion of jawbreakers that weigh more than 0.4 = 60% = 0.6 = \mu _ {\hat p} =p

The formula for the standard deviation of a proportion is \sigma  _{\hat p} =\sqrt{\dfrac{p(1-p)}{n} }

Solving for the standard deviation gives;

\sigma  _{\hat p} =\sqrt{\dfrac{0.6 \cdot (1-0.6)}{800} } = 0.0173

Given that the mean proportion is 0.6, the expected value of jawbreakers that weigh more than 0.4  in the sample of 800 = 800*0.6 = 480

For statistical significance the difference from the mean = 2×\sigma _{\hat p} = 2*0.0173 = 0.0346 the equivalent number of Jaw breakers = 800*0.0346 = 27.7

The z-score of 494 jawbreakers is given as follows;

Z=\dfrac{x-\mu _{\hat p} }{\sigma _{\hat p}  }

Z=\dfrac{494-480 }{0.0173  } = 230.94

Therefore, the z-score more than 2 ×\sigma _{\hat p} which is significant.

8 0
2 years ago
Kathi and Robert Hawn had a pottery stand at the annual Skippack Craft Fair. They sold some of their pottery at the original pri
grigory [225]

Answer:27 pieces were sold at the original price.

63 pieces were sold at the new price

Step-by-step explanation:

Let x represent the number of pieces of pottery that was sold at the original price.

Let y represent the number of pieces of pottery that was sold at the new price.

They sold some of their pottery at the original price of​ $9.50 for each​ piece. This means that the amount that they got from selling x pieces of pottery at the original price would be 9.5x

They later decreased the price of each piece by​ $2. This means that the new price was 9.5 - 2 = $7.5

This means that the amount that they got from selling x pieces of pottery at the new price would be 7.5y

If they sold all 90 pieces and took in ​$729​, then the equations are

x + y = 90

9.5x + 7.5y = 729 - - - - - - - - - -1

Substituting x = 90 - y into equation 1, it becomes

9.5(90 - y) + 7.5y = 729

855 - 9.5y + 7.5y = 729

- 9.5y + 7.5y = 729 - 855

- 2y = - 126

y = - 126/- 2 = 63

Substituting y = 63 into x = 90 - y, it becomes

x = 90 - 63 = 27

6 0
2 years ago
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