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4 years ago

14

It takes 24 megabytes to store 144 songs on a music player. What is the greatest number of songs that 60 megabytes could store?

A. 288 B. 360 C. 540

1 answer:

givi [52]4 years ago

8 0

Option B

We can store a total of 360 songs in 60 megabytes of space.

<u>Solution:</u>

It is given to us that It takes 24 megabytes to store 144 songs on a music player. We have been asked to find out what is the greatest number of songs that 60 megabytes could store.

To solve this question we can use the unitary method.

We will first find out how many songs can be stored in 1 megabyte. This can be done as follows:

$=\frac{144}{24}=6$

Therefore, we can store 6 songs in 1 megabyte.

So, in 60 megabyte we can store the following number of songs:

$=60 \times 6=360$

So, we can store a total of 360 songs in 60 megabytes.

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Is 0.3434434443 a rational number

Kobotan [32]

Answer:

Yes ,0.3434434443 is a rational number.

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3 years ago

Which shows the dimensions of two rectangular prisms that have volumes of 240 ft3 but different surface areas?

bulgar [2K]

Calculate the volume and surface areas of both rectangular prisms for each option.

A:

Rectangular Prism #1
$\sf V=lwh$
$\sf V=(8)(5)(6)=240~ft^3$

$\sf A=2((8)(5)+(6)(5)+(6)(8))$
$\sf A=236~ft^2$

Rectangular Prism #2
$\sf V=lwh$
$\sf V=(5)(6)(8)=240~ft^3$

$\sf A=2((5)(6)+(8)(6)+(8)(5))$
$\sf A=236~ft^2$

This is incorrect, they have the same volume of 240 cubic feet and the same surface areas.

B:

Rectangular Prism #1
$\sf V=lwh$
$\sf V=(8)(5)(6)=240~ft^3$

$\sf A=2(wl+hl+hw)$
$\sf A=2((8)(5)+(6)(5)+(6)(8))$
$\sf A=236~ft^2$

Rectangular Prism #2
$\sf V=lwh$
$\sf V=(10)(8)(3)=240~ft^3$

$\sf A=2(wl+hl+hw)$
$\sf A=2((10)(8)+(3)(8)+(3)(10))$
$\sf A=268~ft^2$

$\boxed{\sf Correct~Answer}$ : Both have the same volume of 240 cubic feet, but different surface areas.

C:

Rectangular Prism #1
$\sf V=lwh$
$\sf V=(15)(8)(2)=240~ft^3$

$\sf A=2(wl+hl+hw)$
$\sf A=2((15)(8)+(2)(8)+(2)(15))$
$\sf A=332~ft^2$

Rectangular Prism #2
$\sf V=lwh$
$\sf V=(6)(4)(15)=360~ft^3$

$\sf A=2(wl+hl+hw)$
$\sf A=2((6)(4)+(15)(4)+(15)(6))$
$\sf A=348~ft^2$

This is incorrect, they have different surface areas, but only Rectangle #1 has a volume of 240 cubic feet.

D:

Rectangular Prism #1
$\sf V=lwh$
$\sf V=(12)(6)(5)=360~ft^3$

$\sf A=2(wl+hl+hw)$
$\sf A=2((12)(6)+(5)(6)+(5)(12))$
$\sf A=324~ft^2$

Rectangular Prism #2
$\sf V=lwh$
$\sf V=(8)(10)(3)=240~ft^3$

$\sf A=2(wl+hl+hw)$
$\sf A=2((8)(10)+(3)(10)+(3)(8))$
$\sf A=268~ft^2$

This is incorrect, they both have different surface areas but only Rectangle #2 has a volume of 240 cubic feet.

5 0

4 years ago

You are going to a bowling alley and have $10 to spend. Renting bowling shoes cost $1.50. Each game cost $2.

MaRussiya [10]

Answer:

6.50

Step-by-step explanation:

10.00
$10.00 - 1.50 - 8.50 - 2.00 = 6.50$

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3 years ago

Find the missing part. L=8, w=4, h=2. Find the diagonal of (d) of the retangular solid

marusya05 [52]

The diagonal of the rectangular solid is $d=2 \sqrt{21}$

Explanation:

The length of the rectangular solid is $l=8$

The width of the rectangular solid is $w=4$

The height of the rectangular solid is $h=2$

We need to determine the diagonal of the rectangular solid.

The diagonal of the rectangular solid can be determined using the formula,

$d=\sqrt{l^{2}+w^{2}+h^{2}}$

Substituting the values $l=8$ , $w=4$ and $h=2$ , we get,

$d=\sqrt{8^{2}+4^{2}+2^{2}}$

Squaring the terms, we get,

$d=\sqrt{64+16+4}$

Adding the terms, we have,

$d=\sqrt{84}$

Simplifying, we have,

$d=2 \sqrt{21}$

Thus, the diagonal of the rectangular solid is $d=2 \sqrt{21}$

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4 years ago

Eva wants to know the probability of tossing “heads” at least three times out of five. She used the random number table to simul

umka21 [38]

To obtain the probability of obtaining heads at least 3 times out of 5 times we recall that in her simulation <span>she assigned odd digits to represent heads and even digits to represent tails. Thus, we count the simulated numbers to see in how many numbers do we have 3 or more odd digits.

Below, the simulated numbers with 3 or more odd digits are bolded.
</span> <span>32766 53855 34591 27732
47406 31022 25144 72662
03087 35521 26658 81704
56212 72345 44019 <span>65311
</span>
We have 6 simulated numbers having 3 or more odd digits.
Therefore, </span><span>P("heads" at least 3 out of 5 times) = 6 / 16 = 3 / 8.</span>

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4 years ago