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3 years ago

8

Consider the function f(x) = 3x2 + 7x + 2.

1 answer:

Slav-nsk [51]3 years ago

8 0

F(x)= 8+ 7x
F(x)=8/7

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5*10 divided by 30 -10

sp2606 [1]

The answer to your question would be -25/3

As decimal -8.333333

4 0

3 years ago

Monica took a survey of her classmates' hair and eye color. The results are in the table below.

katrin2010 [14]

Answer: 0.5

Step-by-step explanation: it should be but wait for other answers

3 0

2 years ago

What is 175% of 14.2

LUCKY_DIMON [66]

175% of 14.2 is saying that

1.75(14.2)=24.9

Any questions please just ask. Thank you.

7 0

3 years ago

Read 2 more answers

1. Points A and B are to be mapped onto a number line according to two equations. The solution to the equation 2/3 A= -24 is the

OleMash [197]

<span>The solution to the equation 2/3 A= -24 is the coordinate of point A.

2/3 A = -24
A = -24 / 2/3
A = -24 * 3/2
A = -72/2
A = -36

The solution to the equation 20=-b/0.5 is the coordinate of point B

20 = -b/0.5
20 * 0.5 = -b
10 = -b
10/-1 = -b/-1
-10 = b

A = -36 and B = -10

The distance between the two letters is 26 units. </span>

4 0

3 years ago

At a local University, it is reported that 81% of students own a wireless device. If 5 students are selected at random, what is

Sphinxa [80]

Answer:

34.87% probability that all 5 have a wireless device

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

81% of students own a wireless device.

This means that $p = 0.81$

If 5 students are selected at random, what is the probability that all 5 have a wireless device?

This is P(X = 5) when n = 5. So

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

$P(X = 5) = C_{5,5}.(0.81)^{5}.(0.19)^{0} = 0.3487$

34.87% probability that all 5 have a wireless device

5 0

3 years ago