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A pack of folders has a length of 5 inches, a width of 12 inches, and a height of 1 inch. The pack of folders will be shipped in

777dan777 [17]

Answer:

A) Each pack of folders has a volume of 60 cubic inches.

B) The box has a volume of about 720 cubic inches

D) If the box help 20 packs of folders, it would have a volume of about 1,200 cubic inches.

Step-by-step explanation:

Verify each statement

A) Each pack of folders has a volume of 60 cubic inches.

The statement is True

Because

The volume of each pack of folders is equal to

$V=(5)(12)(1)=60\ in^{3}$

B) The box has a volume of about 720 cubic inches

The statement is True

Because

The volume of the box is equal to the volume of one pack of folders multiplied by 12

so

$V=(12)60=720\ in^{3}$

C) If the box held 15 packs of folders, it would have a volume of about 1,200 cubic inches

The statement is False

Because

Applying proportion

$\frac{12}{720}\frac{packs}{in^{3}}=\frac{15}{x}\frac{packs}{in^{3}}\\ \\x=720*15/12\\ \\x=900\ in^{3}$

$900\ in^{3}\neq 1,200\ in^{3}$

D) If the box help 20 packs of folders, it would have a volume of about 1,200 cubic inches.

The statement is True

Because

Applying proportion

$\frac{12}{720}\frac{packs}{in^{3}}=\frac{20}{x}\frac{packs}{in^{3}}\\ \\x=720*20/12\\ \\x=1,200\ in^{3}$

$1,200\ in^{3}= 1,200\ in^{3}$

E) Each pack of folders has a volume of 24 cubic inches.

The statement is False

Because

The volume of each pack of folders is equal to

$V=(5)(12)(1)=60\ in^{3}$

7 0

3 years ago

For which of the following compound inequalities is there no solution?

Bess [88]

Answer:

(a) $6m \le - 36$ and $m + 24 > 20$

Step-by-step explanation:

Required

Which of a to d has no solution

(a) $6m \le - 36$ and $m + 24 > 20$

We have: $m + 24 > 20$

Sole for m

$m >20 - 24$

$m >-4$

$6m \le - 36$

Divide by 6

$m \le -6$

So, we have:

$m \le -6$ and $m >-4$

$m \le -6$ implies that: m = -6,-7,-8,-9.....

$m >-4$ implies that m = -3,-2,-1,0.....

Hence, there is no solution

8 0

3 years ago

Suppose I claim that the average monthly income of all students at college is at least $2000. Express H0 and H1 using mathematic

pishuonlain [190]

Answer:

For this case we want to test if the the average monthly income of all students at college is at least $2000. Since the alternative hypothesis can't have an equal sign thne the correct system of hypothesis for this case are:

Null hypothesis (H0): $\mu \geq 2000$

Alternative hypothesis (H1): $\mu$

And in order to test this hypothesis we can use a one sample t or z test in order to verify if the true mean is at least 200 or no

Step-by-step explanation:

For this case we want to test if the the average monthly income of all students at college is at least $2000. Since the alternative hypothesis can't have an equal sign thne the correct system of hypothesis for this case are:

Null hypothesis (H0): $\mu \geq 2000$

Alternative hypothesis (H1): $\mu$

And in order to test this hypothesis we can use a one sample t or z test in order to verify if the true mean is at least 2000 or no

7 0

3 years ago

2

NikAS [45]

Answer:

x=9

Step-by-step explanation:

7x-11 = 4x-2

7x-4x=11-2

3x=9

x=9

3 0

3 years ago

A sprinter can run 279 feet in 9 seconds. Find the sprinter's unit rate of feet per second.

TiliK225 [7]

Easy 31 feet per second take 279 and divide it by 9 and you get 31

5 0

3 years ago