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IgorLugansk [536]
2 years ago
10

PLEASE HELP ASAP I WILL MARK YOU BRAINLIEST!!!

Mathematics
2 answers:
Mashutka [201]2 years ago
7 0

Answer:

Step-by-step explanation:

10400

dangina [55]2 years ago
3 0

Answer:

3200 cm^3

Step-by-step explanation:

1. Calculate the sides of the right triangle to get the height

We know that side c = 26 and the shortest leg is 10 cm (20cm/2) because this right triangle splits the 20cm in two)

10^{2} + b^{2} = 26^{2}   \\100 + b^{2} = 676\\b^{2} = 576 \\b = \sqrt{576} \\b = 24

a = 10, b = 24, c = 26

2. Use the formula for the volume of a square pyramid now that you have the height (which is b)

V = a^{2} \frac{h}{3}  \\\\V = 20^{2} * \frac{24}{3} \\V = (400)(8)\\V = 3200 cm^{3}

V = 3200

I hope this helped! :)

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John and Belinda played a nine holes of golf. John's score was 10 less than two times Belinda's score. If Johns score was 54 str
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Answer:

Belinda's score is 32 strokes.

Step-by-step explanation:

Let the score of John be "x" and Belinda be "y".

It is given that, in the game of golf, John's score was 10 less than two times Belinda's score.

Also, John's score is 54 strokes.

The above equation can be written as ;

x = 2(y) -10

Here, x = 54,

54 = 2(y) -10

64 = 2(y)

y = 32

Thus, Belinda's score is 32 strokes.

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Yana plans to run 1 mile she has run 7/10 of a mile what fraction of a mile does she have left to run
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Read 2 more answers
SAT scores are normed so that, in any year, the mean of the verbal or math test should be 500 and the standard deviation 100. as
vovangra [49]

Answer:

a) P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)

P(Z>1.25)=1-P(Z

b) P(400

P(-1

P(-1

c) z=-0.842

And if we solve for a we got

a=500 -0.842*100=415.8

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

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 SAT scores of a population, and for this case we know the distribution for X is given by:

X \sim N(500,100)  

Where \mu=500 and \sigma=100

We are interested on this probability

P(X>625)

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>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)

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

P(Z>1.25)=1-P(Z

Part b

We are interested on this probability

P(400

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(400

And we can find this probability with this difference:

P(-1

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

P(-1

Part c

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

P(X>a)=0.8   (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 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(Z>-0.842)=0.8

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=-0.842

And if we solve for a we got

a=500 -0.842*100=415.8

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

8 0
3 years ago
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