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serg [7]
3 years ago
11

What is the volume of the prism shown below? O 8 cm O 10 cm O 12 cm3 O 16 cm3

Mathematics
1 answer:
Bogdan [553]3 years ago
3 0

length times width times height

(1¼) (4) (2)

= 10 cm3

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Which term is a perfect square of the root 3x4?
hodyreva [135]

Answer:

85

Step-by-step explanation:

because eight plus five is 85

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I need help with the answer
solmaris [256]

Answer:

Its the last option

y = 1488.33x + 23800 ; $172,600.

Step-by-step explanation:

The slope of the trend line = (median value for 2000 - median value for 1940) / 60

= 89300 / 60

= 1488.33.

So the equation is 1488.33x + 23,800

Estimate for 2040

is 1483.33* 100 + 23800

= 172,600.

5 0
3 years ago
Given that 5 x − 3 y = 32 Find y when x = 4
ollegr [7]

Answer:

<h2><em>y</em><em>=</em><em>-</em><em>4</em></h2>

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<em>X=</em><em>4</em>

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<em>5x - 3y = 32 \\ or \: 5 \times x - 3y = 32 \\ or \: 5 \times 4 - 3y = 32 \\ or \: 20 - 3y = 32 \\ or \:  - 3y = 32 - 20 \\ or \:  - 3y = 12 \\ or \: y = -   \frac{12}{ 3}  \\ y =  - 4</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>

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4 0
3 years ago
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The number of days that homes stay on the market before they sell in Houston is bell-shaped with a mean equal to 56 days. Furthe
antoniya [11.8K]

Answer:

D. 8

Step-by-step explanation:

We have been given that the number of days that homes stay on the market before they sell in Houston is bell-shaped with a mean equal to 56 days. Further, 95 percent of all homes are on the market between 40 and 72 days.

As per empirical rule 95% of the data on bell curve lies between 2 standard deviations of mean.

So we can set an equation as:

72-56=2\sigma or

40-56=-2\sigma

16=2\sigma

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8=\sigma

40-56=-2\sigma

-16=-2\sigma

\frac{-16}{-2}=\frac{-2\sigma}{-2}

8=\sigma

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6 0
2 years ago
Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice earn five candy canes per hour but can
dimaraw [331]

Answer:

  (a) 5 senior, 4 apprentice

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  (c) 7.5 senior, 0 apprentice

Step-by-step explanation:

The problem can be described by two inequalities. On describes the limit on the number of elves in the shop; the other describes the limit on the total payroll. Let x and y represent the number of apprentice and senior elves, respectively. Then the inequalities for the first scenario are ...

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These two inequalities are graphed in the first attachment. They describe a solution space with vertices at ...

  (x, y) = (0, 7.5), (4, 5), (9, 0)

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(a) Santa wants to  maximize the output of trucks, so wants to maximize the function t = 4x +6y.

At the vertices of the solution space, the values of this function are ...

  t(0, 7.5) = 45

  t(4, 5) = 46

  t(9, 0) = 36

Output of trucks is maximized by a workforce of 4 apprentice elves and 5 senior elves.

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(b) The above calculations show 46 trucks per hour can be made, so ...

  46×8 = 368 . . . trucks in an 8-hour shift

__

(c) The new demands change the inequalities to ...

  x + y ≤ 8 . . . . . . number of workers

  7x +8y ≤ 60 . . . total wages (per hour)

The vertices of the feasible region for these condtions are ...

  (x, y) = (0, 7.5), (4, 4), (8, 0)

From above, we know the truck output will be maximized at the vertex (x, y) = (0, 7.5). However, we know we cannot have 7.5 senior elves working in the shop. We can have 7 or 8 elves working.

If the workforce must remain constant, truck output is maximized by a workforce of 7 senior elves.

If the workforce can vary through the shift, truck output is maximized by adding one more senior elf in the shop for half a shift.

Santa should assign 7 senior elves for the entire shift, and 8 senior elves (one more) for half a shift.

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<em>Comment on apprentice elf wages</em>

At 5 candy canes for 4 trucks, apprentice elves produced trucks for a cost of 1.25 candy canes per truck. At 8 candy canes for 6 trucks, senior elves produced trucks for a cost of about 1.33 candy canes per truck. The reason for employing senior elves in the first scenario is that their productivity is 1.5 times that of apprentice elves while their cost per truck is about 1.07 times that of apprentice elves.

After the apprentice elves wages were increased, their cost per truck is 1.75 candy canes per truck, but their productivity hasn't changed. They have essentially priced themselves out of a job, because they are not competitive with senior elves.

5 0
2 years ago
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