a spinner is divided into four equal sections labeled 1 to 4. What is the probability that the spinner will land on 2?

Which answer is equal to the quotient in the expression below? (2x3 - x2) (x2)

serg [7]

(2x3 - x2) / (x2)

= 2x^3/x^2 - x^2/x^2

=2x - 1

answer

2x - 1

4 0

3 years ago

Factorise y^2 - 16 (y squared)

cricket20 [7]

$y^2-16=y^2-4^2=(y-4)(y+4)\\\\\text{used}\ a^2-b^2=(a-b)(a+b)$

8 0

3 years ago

Read 2 more answers

In a survey of 1,532 former UF students, 84% said they were still in love with Tim Tebow. The survey's sampling error was 2%. Us

nalin [4]

Answer:

confidence interval for the proportion of all former UF students who are still in love with Tim Tebow.

(0.79 , 0.89)

Step-by-step explanation:

step 1:-

Given sample survey former UF students n = 1532

84% said they were still in love with Tim Tebow

p = 0.84

The survey sampling error

$= \sqrt{\frac{p(1-p)}{n} }$

Given standard error of proportion = 2% =0.02

Step 2:-

The 99% of z- interval is 2.57

The 99% of confidence intervals are

p ± zₐ S.E (since sampling error of proportion = $\sqrt{\frac{pq}{n} }$

$(0.84 - 2.57X0.02 , 0.84 + 2.57X0.02)$

on simplification , we get

(0.84 - 0.0514 , 0.84 + 0.0514)

(0.79 , 0.89)

conclusion:-

confidence interval for the proportion of all former UF students who are still in love with Tim Tebow.

(0.79 , 0.89)

4 0

3 years ago

Help!!!!!!!!!!!!!!!!!!!!

Ksenya-84 [330]

Answer:

500w + 25 = 200

Step-by-step explanation:

"He withdraws $25 per week..." <------

8 0

2 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

3 years ago