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Korolek [52]
2 years ago
9

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

Mathematics
1 answer:
den301095 [7]2 years ago
3 0

The range of the function y = x2 is all real numbers, y, such that y ≥ 0.

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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 <br> What is the solution of
ANTONII [103]

Answer:

  • Option A

Step-by-step explanation:

<u>Find zeros first:</u>

  • 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)

<u>We have 4 intervals:</u>

  • 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

<u>The solution is:</u>

  • 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&gt;$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
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