Angelina makes 70% of her free throws. What is the probability tha she will make her next two free throws?

Liono4ka [1.6K]

Let's solve this problem using substitution. Given that x-y=8, x = 8 + y.
(Then x^2 = 64 + 16y + y^2)
This other equation is (x-2)^2 + (y-1)^2 = 25.

Easier to substitute 8 + y for x in (x-2)^2:

(8 + y - 2)^2 + (y-1)^2 = 25

(6 + y)^2 + (y-1)^2 = 25
36 + 12y + y^2 + y^2 - 2y + 1 = 25

Re-writing this in descending powers of y:

2y^2 + 10y + 36 + 1 = 25

Then 2y^2 + 10y - 12 = 0

Reduce by division by 2: y^2 + 5y - 6 = 0 = (y+3)(y+2) = 0

Then y=-3 and y=-2. From each of these we get x: x = 8 + y

So x = 8 - 3 = 5 and x = 8 - 2 = 6. There are common solutions.

Try (5, -3) and (6, -2). Do these points satisfy both of the given equations? If they do, you've shown that we have common solutions.

6 0

4 years ago

A random sample of 20 items is selected from a population. When computing a confidence interval for the population mean, what nu

kumpel [21]

Answer:

For this case since the sample size is lower than 30 , n =20<30, is not appropiate use the normal standard distribution to calculate the confidence interval for the mean, and then for this case is better use the t distribution since takes in count the correction factor to approximate the distribution of the true parameter.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

For this case since the sample size is lower than 30 , n =20<30, is not appropiate use the normal standard distribution to calculate the confidence interval for the mean, and then for this case is better use the t distribution since takes in count the correction factor to approximate the distribution of the true parameter.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

8 0

3 years ago

Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all 10! pos

m_a_m_a [10]

Answer:

$P(X=1)=\frac{1}{2}\\P(X=2)=\frac{5}{18}\\P(X=3)=\frac{5}{36}\\P(X=4)=\frac{5}{84}\\P(X=5)=\frac{5}{252}\\P(X=6)=\frac{1}{252} \\P(X>6)=0$

Step-by-step explanation:

Let's define the following event :

X : ''Highest ranking achieved by a woman''

There are five men and five women so the lowest ranking possible is 6, this means that all five men achieved a better ranking than all five women.

In terms of the variable :

X can assume the following values :

$X=1\\X=2\\X=3\\X=4\\X=5\\X=6$

⇒ $P(X=7)=0\\P(X=8)=0\\P(X=9)=0\\P(X=10)=0$

⇒ $P(X>6)=0$

Now let's calculate the probability function for X

$P(X=1)=\frac{(5)(9!)}{(10!)}=0.5=\frac{1}{2}$

We write on the denominator the total ways to arrange ten different persons in a line. This number is 10!

On the numerator we write the total cases in which the first person is a woman ⇒ (5).(9!)

Because they are five different women and when you pick one of then they are 9! ways to arrange the 9 remaining persons

Following this logic, we continue calculating the probability function :

$P(X=2)=\frac{(5)(5)(8!)}{(10!)}=\frac{5}{18}$

Again : Five different ways to choose between the five men in the first position.Then, 5 different ways to choose between the five women.And finally, 8! ways to arrange the remaining persons

$P(X=3)=\frac{(5)(4)(5)(7!)}{(10!)}=\frac{5}{36}$

Here we have : five different ways to choose between five men.Four different ways to choose between the four remaining men.Five different ways to choose between the five women for the third position and finally 7! ways to arrange the remaining persons.

$P(X=4)=\frac{(5)(4)(3)(5)(6!)}{(10!)}=\frac{5}{84}$