A roof has a cross section that is a right triangle. The diagram shows the approximat

WITCHER [35]

Answer:

the answer should be B

Step-by-step explanation:

take the total of people who got the flu(63) and the amount of them who were vaccinated(35) and write it as a fraction. 35/63 in its simplest form is 5/9

8 0

3 years ago

New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today

Ierofanga [76]

Answer:

a) We can find the z score for the value of 225 and we got:

$z = \frac{225-204}{55}=0.382$

And we can find this probability with the complement rule:

$P(z>0.382)=1-P(z$

b) $z = \frac{140-204}{55}=-1.164$

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

$P(z$

c) $P(200$

And we can find this probability with this difference:

$P(-0.072$

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

$P(-0.072$

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

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

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

$a=204 +0.842*55=250.31$

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

$X \sim N(204,55)$

Where $\mu=204$ and $\sigma=55$

We are interested on this probability

$P(X>225)$

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

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

We can find the z score for the value of 225 and we got:

$z = \frac{225-204}{55}=0.382$

And we can find this probability with the complement rule:

$P(z>0.382)=1-P(z$

Part b

$P(X$

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:

$z = \frac{140-204}{55}=-1.164$

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

$P(z$

Part c

$P(200$

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:

$P(200$

And we can find this probability with this difference:

$P(-0.072$

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

$P(-0.072$

Part d

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

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

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

$a=204 +0.842*55=250.31$

5 0

3 years ago

Can you help me please?! :)

UkoKoshka [18]

The answer is A
start by isolating the y variable
the rest of the work is shown here