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2 years ago

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There are 90 fith-grade students going on a feild trip. Each student gives the teacher $9.25 to cover admission to the theater a

nd for lunch. Admission for all the students will cost $315, and each student will get an equal amount to spend on lunch. How much will each fifth grader get to spend on lunch?

1 answer:

garri49 [273]2 years ago

4 0

Answer:

they each get $5.75 for lunch

Step-by-step explanation:

9.25 x 90 students = 832.50

832.50 -315 = 517.50

517.50 / 90 = 5.75

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(b) What portion of the official U.S. poverty threshold (roughly $29,000 for a family of four) is met by the World Bank’s measur

otez555 [7]

World Bank's threshold for poverty is $1.25 per person per day.
For a family of four this gives:
Daily income = (number of people) * threshold
Daily income = 4 * $1.25
Daily income = $5
For a year this is:
Yearly income = (daily income) *365
Yearly income = $5 *365
Yearly income = $1825

U.S. poverty threshold is set at $29000. Portion of threshold can be calculated as:
$Portion\text{ }of\text{ }threshold= \frac{Poverty\text{ }threshold\text{ }(World\text{ }Bank)}{Poverty\text{ }threshold\text{ }(U.S.)} *100\% \\ \\ Portion\text{ }of\text{ }threshold= \frac{1825}{29000} *100\% \\ \\ Portion\text{ }of\text{ }threshold=6.29\%$

The poverty threshold measured by World Bank is 6.29% of the U.S. poverty threshold.

5 0

2 years ago

The data show the average monthly temperatures for two cities over a 6-month period. City 1: {20, 24, 40, 63, 76, 89} City 2: {4

Mrac [35]

Answer:

The MAD of city 2 is <u>less than</u> the MAD for city 1, which means the average monthly temperature of city 2 vary <u>less than</u> the average monthly temperatures for City 1.

Step-by-step explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

$Mean = x1 = \frac{20+24+40+63+76+89}{6}$

$x1 = \frac{312}{6}$

$x1 = 52$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

$20-52 = -32=32\\24-52=-28=28\\40-52=-12=12\\63-52=11\\76-52=24\\89-52=37$

The deviations will be added then. $Mean Absolute Deviation = \frac{32+28+12+11+24+37}{6} \\=\frac{144}{6}\\=24$

So the mean absolute deviation for city 1 is 24 ..

For city 2:

$Mean = x2 = \frac{41+50+58+62+72+83}{6}$

$x2 = \frac{366}{6}$

$x2 = 61$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

$41-61=-20=20\\50-61=-11=11\\58-61=-3=3\\62-61=1\\72-61=11\\83-61=22$

The deviations will be added then. $Mean Absolute Deviation = \frac{20+11+3+1+11+22}{6} \\=\frac{68}{6}\\=11.33$

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is <u>less than</u> the MAD for city 1, which means the average monthly temperature of city 2 vary <u>less than</u> the average monthly temperatures for City 1.

3 0

2 years ago

Read 2 more answers

What is 6÷2(2+1)? explain how did you get the answer

denis-greek [22]

Answer:

$6 \div 2(2 + 1) \\ 6 \div 2(3) \\ 3 \times 3 \\ = 9$

8 0

3 years ago

Read 2 more answers

You are baking brownies for your class. There are 28 total students in your class and you have baked 23 brownies. Right and solv

slavikrds [6]

Answer:

$x = 33$

Step-by-step explanation:

Given

$Students = 28$

$Brownies = 23$

$Brownies = 2\ per\ student$

Required

Determine the number of brownies left

If:

$Brownies = 2\ per\ student$

and there are 28 students

Then:

$Total\ Brownies = 2 * 28$

$Total\ Brownies = 56$

To determine the number of brownies left, we have:

$Total\ Brownies = Brownies\ Made + Brownies\ Left$

$56 = 23 + x$

Reorder

$x + 23 = 56$

Collect Like Terms

$x = 56 - 23$

$x = 33$

<em>Hence, there are 33 brownies left</em>

6 0

3 years ago

Does this graph show a function? explain how you know

AVprozaik [17]

Answer:

C

Step-by-step explanation:

The vertical line test is basically just drawing a vertical line and seeing if the line intersects the graph more than once. If it does, then it is not a function, if it doesn't than it is a function.

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3 years ago

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