All categories

3 years ago

Factor out the greatest common factor -5x^2 - 5x^3 - 15x^4

1 answer:

5 0

Our answer is $-5x^{2} (3x^{2} +x+1)$

Our greatest common factor, or GCF, is $-5x^{2}$ , so when we factor is out, our divide each term by $-5x^{2}$ , we get our answer. To check this, we can multiply the $-5x^{2}$ back across the parenthesis and see if we get the same result as our initial question.

Hope this helped! :))

You might be interested in

Find the length of x exactly. Brainly!!

NNADVOKAT [17]

Answer:

x = 7

Step-by-step explanation:

We can tell by looking at this triangle that this is a <u>45 45 90 triangle.</u>

In this situation, the hypotenuse is $x\sqrt{2}$ and the legs are both $x$ .So if we look at the hypotenuse, we can tell that x, or the leg of the triangle, is 7.

45 45 90 triangles: This is a special triangle in which the angles of the triangle are 45 45 and 90 degrees. This means that the hypotenuse will be $x\sqrt{2}$ and the legs of the triangle will be $x$ . You can usually use the Pythagorean theorem to solve the missing sides.

8 0

3 years ago

**WILL MARK BRAINLIEST** PLEASE ANSWER!! please ??

geniusboy [140]

Answer:

Port eagle has the higher median

Step-by-step explanation:

That's all I will give you.

Why?

Because the median is the average which would be the middle range between two points.

7 0

3 years ago

Elsa buys 27 quarts of punch for a party. Each serving is 8 fluid ounces. If the guests drink 100 servings of punch, how many fl

a_sh-v [17]

Answer:

764 US Fluid Ounce

Step-by-step explanation:

Refrence Sheet Used: https://media.studyisland.com/pics/FL8thgradereferencesheet.html

27 US Liquid Quarts = 864 US Fluid Ounces, - 100 US Fluid Ounces = 764

4 0

3 years ago

Bryce is making a model building. He raises the walls of the building by 2 centimeters, five times.

photoshop1234 [79]

Answer:

10 cm!

I think...it's really basic math so...maybe?

7 0

3 years ago

Read 2 more answers

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago