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nalin [4]
3 years ago
11

Need help anything would help at this point !!!? Extra points

Mathematics
1 answer:
Allisa [31]3 years ago
4 0

Answer:

here,radius =2 in

now use formula pie (r)^2 where pie is 3.14

=pie *(r)^2

=3.14*(2in.)^2

=3.14*4in.^2

12.56in.^2

Step-by-step explanation:

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Larry Lazy purchased a one year membership at a local fitness center at the beginning of the year. It cost him $150. He goes twi
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$4.05 per visit
if he continued going to the gym twice a week all year would be $2.88 per visit.
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3 years ago
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Kim picked 250 pounds of pears yesterday and 287 pounds of pears today. What is the percentage increase of
xxMikexx [17]

Given:

Kim picked 250 pounds of pears yesterday and 287 pounds of pears today.

To find:

The percentage increase of  amount of pears picked by Kim.

Solution:

We know that,

\%\text{ increase}=\dfrac{\text{New value - Initial value}}{\text{Initial value}}\times 100

\%\text{ increase}=\dfrac{287-250}{250}\times 100

\%\text{ increase}=\dfrac{37}{25}\times 10

\%\text{ increase}=\dfrac{370}{25}

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7 0
2 years ago
Please please help. Marking Brilliant for the correct answer
NISA [10]

Answer:

x = 19

Step-by-step explanation:

(whole secant) x (external part) = (tangent)^2

(x-7+4) * 4 = 8^2

(x-3)*4 = 64

Divide each side by 4

x-3 =16

Add 3 to each side

x-3+3 = 16+3

x = 19

5 0
3 years ago
Evaluate 53 – 2m for m = 3.
irakobra [83]
53 - 2(3)
53 - 6
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3 0
3 years ago
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The distribution of a sample of the outside diameters of PVC pipes approximates a normal distribution. The mean is 14.0 inches,
lakkis [162]

Answer:

A. 13.9 and 14.1 inches

See explanation below.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the outside diameters of a population, and for this case we know the distribution for X is given by:

X \sim N(14,0.1)  

Where \mu=14 and \sigma=0.1

If we want the middle 68% of the data we need to have on the tails 16% on each one

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.84   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.16 of the area on the left and 0.84 of the area on the right it's z=-0.994. On this case P(Z<-0.994)=0.16 and P(z>-0.994)=0.84

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.994

And if we solve for a we got

a=14 -0.994*0.1=13.9

And since the distribution is symmetrical for the upper limit we can use z = 0.994 and we have:

z=0.994

And if we solve for a we got

a=14 +0.994*0.1=14.1

So the correct answer for this case would be:

A. 13.9 and 14.1 inches

3 0
3 years ago
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