If f(x) = 3x + 5 and g(x) = –2x2 – 5x + 3, find (f + g)(x).

Test scores of the student in a school are normally distributed mean 85 standard deviation 3 points. What's the probability that

Mrrafil [7]

Answer:

The probability that a random selected student score is greater than 76 is $\\ P(x>76) = 0.99865$ .

Step-by-step explanation:

The Normally distributed data are described by the normal distribution. This distribution is determined by two parameters, the population mean $\\ \mu$ and the population standard deviation $\\ \sigma$ .

To determine probabilities for the normal distribution, we can use the standard normal distribution, whose parameters' values are $\\ \mu = 0$ and $\\ \sigma = 1$ . However, we need to "transform" the raw score, in this case x = 76, to a z-score. To achieve this we use the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

And for the latter, we have all the required information to obtain z. With this, we obtain a value that represent the distance from the population mean in standard deviations units.

<h3>The probability that a randomly selected student score is greater than 76</h3>

To obtain this probability, we can proceed as follows:

First: obtain the z-score for the raw score x = 76.

We know that:

$\\ \mu = 85$

$\\ \sigma = 3$

$\\ x = 76$

From equation [1], we have:

$\\ z = \frac{76 - 85}{3}$

Then

$\\ z = \frac{-9}{3}$

$\\ z = -3$

Second: Interpretation of the previous result.

In this case, the value is three (3) standard deviations below the population mean. In other words, the standard value for x = 76 is z = -3. So, we need to find P(x>76) or P(x>-3).

With this value of $\\ z = -3$ , we can obtain this probability consulting the cumulative standard normal distribution, available in any Statistics book or on the internet.

Third: Determination of the probability P(x>76) or P(x>-3).

Most of the time, the values for the cumulative standard normal distribution are for positive values of z. Fortunately, since the normal distributions are symmetrical, we can find the probability of a negative z having into account that (for this case):

$\\ P(z>-3) = 1 - P(z>3) = P(z$

Then

Consulting a cumulative standard normal table, we have that the cumulative probability for a value below than three (3) standard deviations is:

$\\ P(z$

Thus, "the probability that a random selected student score is greater than 76" for this case (that is, $\\ \mu = 85$ and $\\ \sigma = 3$ ) is $\\ P(x>76) = P(z>-3) = P(z$ .

As a conclusion, more than 99.865% of the values of this distribution are above (greater than) x = 76.

We can see below a graph showing this probability.

As a complement note, we can also say that:

$\\ P(z3)$

Which is the case for the probability below z = -3 [P(z<-3)], a very low probability (and a very small area at the left of the distribution).

5 0

3 years ago

The average annual cost of tuition, room, and board for

lorasvet [3.4K]

Answer:

Step-by-step explanation: solve it plz

3 0

3 years ago

A=4pi r^{2} solve for r

lara [203]

A = 4πr² Divide both sides by 4π
A / 4π = r² Find the square root of both sides
√(A / 4π) = √(r²) Cancel out the square with the square root
√(A / 4π) = r Switch the sides to make it easier to read
r = √(A / 4π)

5 0

3 years ago

I NEED HELP WITH BOTH I WILL MARK THE BRAINIEST

Dennis_Churaev [7]

First pic: x=27
second pic: x=15

5 0

2 years ago

What is the product of (3p-7)(2p²-3p-4)

garik1379 [7]

Answer:

Choice C is the correct answer.

Step-by-step explanation:

Given expression is

(3p-7)(2p²-3p-4)

we have to find the product of linear expression to quardatic expression.

Firstly, multiply 3p to quardatic expression and -7 to quardatic expression and add.

3p(2p²-3p-4)-7(2p²-3p-4)

Multiply 3p to each term of quardatic expression and -7 to each term of quardatic expression and add all terms:

3p(2p²)+3p(-3p)+3p(-4)-7(2p²)-7(-3p)-7(-4)

6p³-9p²-12p-14p²+21p+28

add like terms

6p³+(-9-14)p²+p(-12+21)+28

6p³-23p²+9p+28 which is the correct answer.

3 0

3 years ago