The answer, I’ve been stuck on this and stressed out.

What is the formula for calculating the area of a circle.

Tom [10]

Answer:

$A=\pi\times r^2$

Step-by-step explanation:

In order to find the area of a circle we multiply π by the radius, and to find the radius of a circle we have to look at the line that stops directly in the middle of the circle.

$\pi\times r$

Once we then have the radius, we simply multiply the radius, (with an exponent of 2) by π and then we will have our answer. Of course, round if needed.

$A=\pi\times r^2$

Hope this helps.

6 0

2 years ago

Issac made a mistake in his checkbook. he wrote a check for $8.98 to rent a video game but mistakenly recorded it in his checkbo

Naya [18.7K]

Debit is where you have to pay extra when you use it. Credit doesn't!!!!!

3 0

3 years ago

Risk taking is an important part of investing. In order to make suitable investment decisions on behalf of their customers, port

GrogVix [38]

Answer:

$a=49.5 -1.28*15=30.3$

So the value of height that separates the bottom 10% of data from the top 90% is 30.3.

If the score is lower than 30.3 we consider this score as risk averse

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

$X \sim N(49.5,15)$

Where $\mu=49.5$ and $\sigma=15$

We are interested in the bottom 10% of the data.

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

$P(X>a)=0.9$ (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 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9

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:

$z=-1.28$

And if we solve for a we got

$a=49.5 -1.28*15=30.3$

So the value of height that separates the bottom 10% of data from the top 90% is 30.3.

If the score is lower than 30.3 we consider this score as risk averse

7 0

3 years ago

Given: f(x)=x-7and h(x) = 2x+3

Musya8 [376]

I have no idea math is so freaking hard x

5 0

3 years ago

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

7 0

3 years ago