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nika2105 [10]
3 years ago
7

Which unit rate is the lowest price per ounce? Choice A: 10 ounces of chocolate chips for $1.66 Choice B: 20 ounces of chocolate

chips for $3.32
A. Choice A
B. Choice B
C. The unit rates are equal
D. The unit rates cannot be determined
Mathematics
2 answers:
Goryan [66]3 years ago
8 0
C.

To find out the rates, you must multiply or divied the ounces and price of on bag of chips to equal the other.

You can divied the 20 ounces into 10 ounces, so you must also divide 3.32 by.
The equals 1.66
Tcecarenko [31]3 years ago
7 0

Answer:  The correct option is

(C) The unit rates are equal.

Step-by-step explanation:  We are given to choose the lowest price per ounce from the following choices :

Choice A: 10 ounces of chocolate chips for $1.66.

Choice B: 20 ounces of chocolate chips for $3.32.

<u>For Choice A :</u>

Price of 10 ounces of chocolate chips = $1.66.

So, the price of 1 ounce of chocolate chips will be

\$\dfrac{1.66}{10}=\$0.166.

<u>For Choice B :</u>

Price of 20 ounces of chocolate chips = $3.32.

So, the price of 1 ounce of chocolate chips will be

\$\dfrac{3.32}{20}=\$0.166.

Therefore, the price per ounce of both the choices A and B are equal.

Thus, the unit rates are equal.

Option (C) is CORRECT.

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LuckyWell [14K]

Answer:

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Step-by-step explanation:

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3 years ago
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean \mu and standard deviation \sigma, we have these following probabilities

Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827

Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545

Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545

Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973

-----------

Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0
3 years ago
Which order pair is a solution of this equation. 5x+y=-25
crimeas [40]

Answer:

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RoseWind [281]
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1: -15   So the change in elevation per  min. is 15

So, the 10 min section of the walk the change of elevation was greater.      Hope that helps! if you need me to explain something just ask!
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8 0
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