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3 years ago

Smiths bike shop rents bikes for $10

Mathematics

1 answer:

Trava [24]3 years ago

3 0

10+(4x3)= $22 (How much she rent the bike for)

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SOMEONE PLEASE HELP WILL MARK BRAINLIEST

Mashutka [201]

Radical form for 112 would be 4{7 and then I think you can just search up how to reduce the 4 to 7 i don’t have the symbol btw

4 0

3 years ago

The resting heart rate of an adult is usually about 60 beats per minute. Use your completed table to find approximately how many

Dvinal [7]

60x=1,000,000
----- ---------------
60 60
X=16,666 2/3 or
X=16,666.6667 it would take approximately 16,667 minutes for the average resting heart to beat 1,000,000 times.

3 0

3 years ago

Which one is greater 10.05 or 10.005

Blababa [14]

10.05 = 10 5/100 = 10 1/20
10.005 = 10 5/1000 = 10 1/200

therefore, 10.05 is greater then 10.005

8 0

3 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

Aleksandr-060686 [28]

Answer:

One difference would be the slope of the lines. One similarity would be where the lines cross the y-axis.

Step-by-step explanation:

This can be seen when we use the equation y=mx+b. m stands for the slope of the line and b stands for where the line is going to cross the y-axis.

I hope this helps.

-Cam ♡

7 0

3 years ago

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