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2 years ago

Find the exact circumference of a circle with the given radius.

Mathematics

2 answers:

kozerog [31]2 years ago

8 0

We know, Circumference = 2π * x/2 = <span>π x

In short, Your Answer would be: Option B

Hope this helps!</span>

vladimir1956 [14]2 years ago

8 0

The answer would be : <span>B) xπ miles.

Hope this helps !

Photon</span>

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Malik drinks 48 ounces of water each day. How many ounces of water will he drink in 16 days? 64 192 336 768

aleksandrvk [35]

Answer: 768 ounces

Step-by-step explanation: ounces of water 48 days 16 multiply 48 x 16 you get 768

6 0

11 months ago

How do you graph the quadratic

iragen [17]

To factor this quadratic function, there are many ways that you can do this, the most common approach used to graph anything would be to make a table of values. Basically, make a chart with x coordinates on one side and y coordinates on the other side. Pick any values for your x coordinates and then find the correct y coordinate that results when you place that specific x coordinate into the function. After getting many points, plot those points on the grid and it should give you a parabola.

Another way to graph the function is to find the x intercepts, you can do this by setting the function to zero, and factoring the trinomial, using the appropriate method, then find the x intercepts

y = 2x^2 - 8x + 6

0 = (2x^2 - 6x - 2x + 6)

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0 = (2x - 2)(x - 3).

3 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.