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3 years ago

Using the ratio of perfect squares method, what is square root of 22 rounded to the nearest hundredth?

1 answer:

5 0

$4+ \frac{22-16}{25-16} = \\\\ 4 + \frac{6}{9} =\\\\4 + \frac{2}{3} 
=\\\\ \boxed{\bf{4.67}}$
Using the ratio of perfect squares method, the square root of 22 rounded to the nearest hundredth is 4.67

Hope this helps!! If you have any other questions or would like further explanation just let me know!! :)

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Austins movie collections contains 9 movies, including 3 comedies. What is the probability that a randomly selected movie will b

storchak [24]

Answer:

There is a 3 in 9 chance

Step-by-step explanation:

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3 years ago

HELP NEEDED ASAP MATHS PLEASE!!!

BabaBlast [244]

You plug in the values for x in each respective equation.

a) f(3) = 3 + 1 which equals: 4.
so, f(3) = 4

b) g(3) = (3)^2 - 3 which equals 9 - 3, which equals 6.
so, g(3) = 6.

hope this helps! (:

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3 years ago

David has 10 yd. Of material that he wiil cut into 15in. Thick mats. How many mats can David make?

Marysya12 [62]

<h2>Answer: He can make <u><em>24</em></u> mats.</h2>

Step-by-step explanation:

10 yards is equal to 360 inches. If you divided 360 inches by 15 inches, the answer will be 24. So the answer is 24 mats.

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2 years ago

If f(x) = –3x – 2, what is f(–5)?

Step2247 [10]

F(x) = -3x - 2
f(-5) = -3(-5) -2 = 15 - 2 = 13

3 0

3 years ago

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) $P(400$

$P(-1$

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