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4 years ago

K(t)=10t−19 k=-7 k=-7= Please help

Mathematics

1 answer:

igomit [66]4 years ago

6 0

Answer:

t = 6/5

Step-by-step explanation:

Step 1: Define

k(t) = 10t - 19

k(t) = -7

Step 2: Substitute and Evaluate

-7 = 10t - 19

12 = 10t

t = 6/5

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A=556

Step-by-step explanation:

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4 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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4 years ago

Solve the system. <br> 3x+2y=-2 6x-Y=-6

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So we assume that x=x and y=y is true for both situations so

3x+2y=-2
6x-y=6

6x-y=6
add y to both sides
6x=6+y
subtract 6 from both sides
6x-6=y
subsitute (6x-6) for y in first equation
3x+2(6x-6)=-2
distribute
3x+12x-12=-2
add like terms
15x-12=-2
add 12 to both sides
15x=10
divide both sides by 15
x=2/3
subsitute
3x+2y=-2
3(2/3)+2y=-2
6/3+2y=-2
2+2y=-2
subtract 2 from both sides
2y=-4
divide both sides by 2
y=-2

x=2/3
y=-2
solution in (x,y) form
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3 years ago

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3 years ago

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What's the question? :)

Step-by-step explanation:

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3 years ago