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Neko [114]
3 years ago
14

Find the area of this figure. Round your answer to the nearest hundredth. Use 3.14 to approximate

Mathematics
1 answer:
arlik [135]3 years ago
3 0

Answer:

26.13

Step-by-step explanation:

to find the area of a circle, it's pi*r squared:

pi*3squared=9pi

it's a semi-circle so you divide it by two=4.5pi

they are telling us to use 3.14 instead of pi, so it would be:

4.5*3.14=14.13

now we find the area of the triangle. And we use the formula, (height*base)/2:

6*4/2=12

now we add the sums together, which is:

12+14.13=26.13

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Drupady [299]
I believe it’s C. 3/2
5 0
2 years ago
Given f(2) = 3, f'(2) = 4, 8(2) = -2, and g'(2) = -4. Find<br> h'(2) if h(x) = f(x) x g(x).
Shtirlitz [24]
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6 0
3 years ago
In a recent year, the ACT scores for the math portion of the test were normally distributed, with a mean of 21.1 and a standard
Natali5045456 [20]

Answer:

a) P(X

And we can find this probability using the normal standard table or excel:

P(z

b) P(19

And we can find this probability with this difference:

P(-0.396

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-0.396

c) P(X>26)=P(\frac{X-\mu}{\sigma}>\frac{26-\mu}{\sigma})=P(Z>\frac{26-21.1}{5.3})=P(z>0.925)

And we can find this probability using the complement rule and the normal standard table or excel:

P(z>0.925)=1- P(Z

d) We can consider unusual events values above or below 2 deviations from the mean

Lower = \mu -2*\sigma = 21.1 -2*5.3 =10.5

A value below 10.5 can be consider as unusual

Upper = \mu +2*\sigma = 21.1 +2*5.3 =31.7

A value abovr 31.7 can be consider as unusual

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".  

Part a

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

X \sim N(21.1,5.3)  

Where \mu=21.1 and \sigma=5.3

We are interested on this probability

P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X

And we can find this probability using the normal standard table or excel:

P(z

Part b

P(19

And we can find this probability with this difference:

P(-0.396

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-0.396

Part c

P(X>26)=P(\frac{X-\mu}{\sigma}>\frac{26-\mu}{\sigma})=P(Z>\frac{26-21.1}{5.3})=P(z>0.925)

And we can find this probability using the complement rule and the normal standard table or excel:

P(z>0.925)=1- P(Z

Part d

We can consider unusual events values above or below 2 deviations from the mean

Lower = \mu -2*\sigma = 21.1 -2*5.3 =10.5

A value below 10.5 can be consider as unusual

Upper = \mu +2*\sigma = 21.1 +2*5.3 =31.7

A value abovr 31.7 can be consider as unusual

6 0
3 years ago
Use the distributive property to remove the parentheses.<br> 3(6-y)
Burka [1]

Answer:

18-3y.

Step-by-step explanation:

3 x 6 = 18

3 x -1y= -3y.

3 0
2 years ago
Read 2 more answers
What is 26.99 round to the nearest tenth
Arte-miy333 [17]
Salutations!

To round to the nearest tenth, you need know whether the number next to tenth place is greater than 5 or lesser than 5. If its greater than 5, you need to round up. If its lesser than 5, you need to round down. Now, in the number 26.99 9 is in the tenth place. The number next to 9 is greater than 5, so just add on to the ones place, making the tenth and hundredth place 0. Zero has no value so whether you add the zero or not, it does not matter.

<span>26.99 round to the nearest tenth </span>≈ 27

Hope I helped (:

Have a great day!
5 0
3 years ago
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