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3 years ago

3t/2+7=4t-3 please answer asap and step by step please

2 answers:

7 0

<span>t in (-oo:+oo)

(3*t)/2+7 = 4*t-3 // - 4*t-3

(3*t)/2-(4*t)+3+7 = 0

(3*t)/2-4*t+3+7 = 0

10-5/2*t = 0 // - 10

-5/2*t = -10 // : -5/2

t = -10/(-5/2)

t = 4

t = 4 is the answer

hope it helps~

</span>

suter [353]3 years ago

6 0

3t/2+7= 4t-3

Try to get the t all by itself.

Combine like terms.

3t/9= 4t-3

Multiply 3t on both sides.

9= 12t-3

Add the three on both sides.

12=12t

Divide by 12 into both sides.

t=1

I hope this helps!
~cupcake

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A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

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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:

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

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

-----------

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

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

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

-----

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.