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2 years ago

Can someone solve this problem and explain to me how you got it

Mathematics

1 answer:

Lady_Fox [76]2 years ago

8 0

Answer:

i do not know i am in 7th grade

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Todd bought a game for $36 with a coupon for 25% off. How much would it have cost without the coupon?

kotykmax [81]

Answer:

Step-by-step explanation:

x*0.75=36

x=36/0.75=48 $

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2 years ago

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A store is having a sale for 30% off. You want to but a shirt that is originally $37. How much will the shirt cost AFTER they ta

Leokris [45]

Answer:

A. 25.90

Step-by-step explanation:

37 x .3 = 11.1

37 - 11.1 = 25.9

Feel free to let me know if you need more help. :)

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2 years ago

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PLEASE HELP!

shusha [124]

For -1:

y = x + 6

y = 5

( -1, 5)

For 0:

y = x + 6

y = 6

(0, 6)

For 2:

y = x + 6

y = 8

(2,8)

For 4:

y = x + 6

y = 10

(4,10)

8 0

2 years ago

Will bought 36 balloons and divided them evenly into 6 groups. How many balloons are in each group?

fredd [130]

Answer:

Step-by-step explanation:

Since the 36 balloons are being divided evenly into 6 groups, you divide the 36 by 6. Once you do this, you are left with 6.

3 0

3 years ago

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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

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

Can someone solve this problem and explain to me how you got it​

Can someone solve this problem and explain to me how you got it