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2 years ago

Which expressions are in their simplest form? Check all that apply.

Mathematics

2 answers:

Tom [10]2 years ago

6 0

Answer:

The first one is the only one in simplest form.

monitta2 years ago

5 0

Answer: this is answerrrr

peace

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A certain recipe ask for 6/7 cup of granola. How much is needed to make 4/5 of a recipe?

Flauer [41]

Well the answer would be 24/35. hope that helped

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2 years ago

One cell phone carrier charges 26.50 a month plus 0.15

Gwar [14]

So what do you need us to do for you?

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2 years ago

Whats three billion plus five trillion lol

Ira Lisetskai [31]

Answer:

5,003,000,000,000

Step-by-step explanation:

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2 years ago

Read 2 more answers

A total of 133 million square miles of the Earth's surface are covered in water. The ratio of the area covered in water to the a

Anni [7]

Answer:

57 million square miles

Step-by-step explanation:

Let the area of water is w and area of land is l.

<u>Their ratio is:</u>

w/l = 7/3

133 / l = 7/3

Find the value of l

l = 133*3/7
l = 57 million square miles

6 0

1 year 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