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3 years ago

13

Patricia is riding a bus to her grandmother's, 66 miles away. On a map with a scale of 2 inches equals 12 miles, what is the dis

tance between her house and her grandmother's house?

2 answers:

Genrish500 [490]3 years ago

7 0

The answer is 5.5 because 66/12= 5.5, hope this helps..

Vedmedyk [2.9K]3 years ago

5 0

Answer:

the answer would be 11 because i did it on usatestprep and the answer was 11

Step-by-step explanation:

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A rectangular bedroom is 10 feet by 12 feet. How many square titles that are 4 inches on each side must be used completely to co

Lilit [14]

Answer:
1,080
Why:
First. you have to turn the feet into inches which would make the room 120in. by 144in. and the area of the room would be 17,280 square inches.
Then you have to find the area of the square tiles, and since the sides of the square are the same length , 4x4=16
Last, you have to divide 17,280 by 16 which would give you 1080 tiles

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3 years ago

Which of the following describes the correct process for solving the equation 3x+6=21 and arrives at the correct solution?

My name is Ann [436]

The correct solution is C

Subtract 6 from both sides of the equation
3x + 6 - 6 = 21 - 6
3x = 15

Then divide both sides by 3

$\frac{3x}{3} = \frac{15}{3}$

x = 5

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3 years ago

An urn contains 8 red chips, 10 green chips, and 2 white chips. A chip is drawn and replaced, and then a second chip is drawn.

Harlamova29_29 [7]

Answer:

(A) 0.04

(B) 0.25

(C) 0.40

Step-by-step explanation:

Let R = drawing a red chips, G = drawing green chips and W = drawing white chips.

Given:

R = 8, G = 10 and W = 2.

Total number of chips = 8 + 10 + 2 = 20

$P(R) = \frac{8}{20}=\frac{2}{5}\\P(G)= \frac{10}{20}=\frac{1}{2}\\P(W)= \frac{2}{20}=\frac{1}{10}$

As the chips are replaced after drawing the probability of selecting the second chip is independent of the probability of selecting the first chip.

(A)

Compute the probability of selecting a white chip on the first and a red on the second as follows:

$P(1^{st}\ white\ chip, 2^{nd}\ red\ chip)=P(W)\times P(R)\\=\frac{1}{10}\times \frac{2}{5}\\ =\frac{1}{25} \\=0.04$

Thus, the probability of selecting a white chip on the first and a red on the second is 0.04.

(B)

Compute the probability of selecting 2 green chips:

$P(2\ Green\ chips)=P(G)\times P(G)\\=\frac{1}{2} \times\frac{1}{2}\\ =\frac{1}{4}\\ =0.25$

Thus, the probability of selecting 2 green chips is 0.25.

(C)

Compute the conditional probability of selecting a red chip given the first chip drawn was white as follows:

$P(2^{nd}\ red\ chip|1^{st}\ white\ chip)=\frac{P(2^{nd}\ red\ chip\ \cap 1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\=\frac{P(2^{nd}\ red\ chip)P(1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\= P(R)\\=\frac{2}{5}\\=0.40$

Thus, the probability of selecting a red chip given the first chip drawn was white is 0.40.

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dem82 [27]

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3 years ago

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Gnoma [55]

Hope this helps ! Answer =1/2f19 and picture

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3 years ago