What is the range of the function y = x2?

Write the set of in roster form if M is the set of integers that are greater that -1 and less than 4?

olchik [2.2K]

If I could understand what you meant my answer is: Set M consisting of 0,1,2 and 3 whose power set has 16 individual sets :)

8 0

3 years ago

PLEASE HURRY WILL GIVE BRAINLIEST What is the solution of

ANTONII [103]

Answer:

Option A

Step-by-step explanation:

Find zeros first:

x² + x - 6 = x² + 3x - 2x - 6 = x(x + 3) - 2(x + 3) = (x - 2)(x + 3)
x = 2, x = -3, x = 7 (but x ≠ 7 as no division by zero)

We have 4 intervals:

x ≤ - 3, three negatives, the result is negative or zero
-3 ≤ x ≤ 2, two negatives and one positive, the result is positive or zero
2 ≤ x < 7, two positives and one negative, the result is negative or zero
x > 7, three positives, the result is positive or zero

The solution is:

x ≤ - 3 or 2 ≤ x < 7

Correct choice is A

8 0

2 years ago

The number is odd.

77julia77 [94]

Answer:

Wow thats pretty hard the only things i could guess was the first one could be 18+ 18=36 and then there 6X6=36. but neither of those are greater than 50. And both of them are less than 75.

Step-by-step explanation:

Is this for a class??

7 0

3 years ago

Which of the following represents a Type I error for the null and alternative hypotheses H0:μ≤$3,200 and Ha:μ>$3,200, where μ

Nuetrik [128]

Answer:

Option a) Type I error would occur if we reject null hypothesis and conclude that the average amount is greater than $3,200 when in fact the average amount is $3,200 or less.

Step-by-step explanation:

We are given the following information in the question:

$H_{0}: \mu \leq \$3,200\\H_A: \mu > \$3,200$

where μ is the average amount of money in a savings account for a person aged 30 to 40.

Type I error:

Type I error is also known as a “false positive” and is the error of rejecting a null hypothesis when it is actually true.
In other words, this is the error of accepting an alternative hypothesis when the results can be attributed by null hypothesis.
A type I error occurs during the hypothesis testing process when a null hypothesis is rejected, even though it is correct and should not be rejected.

Thus, in the above hypothesis type error will occur when we reject the null hypothesis even when it is true.

Option a) Type I error would occur if we reject null hypothesis and conclude that the average amount is greater than $3,200 when in fact the average amount is $3,200 or less.

3 0

3 years ago

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used 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.

41% of U.S. adults have very little confidence in newspapers.

This means that $p = 0.41$

You randomly select 10 U.S. adults.

This means that $n = 10$

(a) exactly five

This is $P(X = 5)$ . So

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

$P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087$

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480$

$P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125$

$P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019$

$P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001$

Then

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834$

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

(c) less than four.

This is:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051$

$P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355$

$P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111$

$P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058$

So

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575$

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0

2 years ago

What is the range of the function y = x2? ​

What is the range of the function y = x2?