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shtirl [24]
3 years ago
10

What is 120% of $590​

Mathematics
2 answers:
Alecsey [184]3 years ago
6 0

Answer:

$708

Step-by-step explanation:

120% of $590 is $708

Nana76 [90]3 years ago
3 0

Answer:

Tip

$708.00

Total

$1,298.00

Step-by-step explanation:

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Match the verbal description with the part of the graph it describes.
lara31 [8.8K]

Answer:

1F

2C

3A

4E

5D

6B

Step-by-step explanation:

Good luck!

8 0
2 years ago
The diameter of a circle 8 inches whats is the area?
pshichka [43]

Answer:

25.12inch^2

Step-by-step explanation:

Area= pi r^2 = 3.14 x 8 =25.12inch^2

6 0
3 years ago
Read 2 more answers
X^2+0x-1=0 what is the number of real solutions<br> A.0<br> B.1<br> C.2
mash [69]
X + 1 hope this hepls
8 0
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%.

4 0
3 years ago
Danny carries 2 gallons of lemonade for distributing in class. How many cups of lemonade has he carried?
SSSSS [86.1K]

Answer:

32

Step-by-step explanation:

multiply the volume value by 16

6 0
2 years ago
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