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2 years ago

27/8+11/3 please help

Mathematics

1 answer:

a_sh-v [17]2 years ago

8 0

Answer:

The answer is 7 1/24

Step-by-step explanation:

27/8 + 11/3

I multipled 27/8 by 3/3 (needed the 8 to become 24)

I multiplied 11/3 by 8/8 (needed the 3 to become 24)

We get:

81/24 + 88/24

Then you add

169/24

Simplify

7 1/24

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Step-by-step explanation:

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2 years ago

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I don’t know what to do with the 3 feet above the ground

Slav-nsk [51]

Answer:

3 seconds

Step-by-step explanation:

You want to set h(t)=3, then take the root such that t≠0. You would have:

3=3+48t−16t^2

Let's solve your equation step-by-step.

3=3+48t−16t2

Step 1: Simplify both sides of the equation.

3=−16t2+48t+3

Step 2: Subtract -16t^2+48t+3 from both sides.

3−(−16t2+48t+3)=−16t2+48t+3−(−16t2+48t+3)

16t2−48t=0

Step 3: Factor left side of equation.

16t(t−3)=0

Step 4: Set factors equal to 0.

16t=0 or t−3=0

t=0 or t=3

3 makes more sense than 0 because the ball can't have been in the air for 0 seconds.

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2 years ago

WILL MARK BRAINLIEST!!!!PLZ HELP!!!! An expression is shown below: 3pf2 − 21p2f + 6pf − 42p2 Part A: Rewrite the expression by f

Lesechka [4]

Answer:

See below.

Step-by-step explanation:

So, we have:

$3pf^2-21p^2f+6pf-42p^2$

PART A:

Find the GCF. Notice that we have a 3 in every term and a p in every term. Thus, the GCF is 3p:

$3p(f^2-7pf+2f-14p)$

This is the most we can do.

PART B:

Continue from where we left off. Factor the entire expression:

$3p(f^2-7pf+2f-14p)\\3p(f(f-7p)+2(f-7p))\\3p((f+2)(f-7p))$

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2 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%.

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3 years ago