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3 years ago

Expand the following question 3(3x-2)

Mathematics

2 answers:

juin [17]3 years ago

6 0

Times 3x and -2 by 3.
3 x 3x = 9x
3 x -2 = -6
Therefore, 3(3x-2) = 9x-6

marissa [1.9K]3 years ago

5 0

3(3x - 2)

Distribute (multiply) 2 inside the parentheses.

3 * 3x and 3 * -2

9x - 6

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28 1/2% covert the percent to a decimal

rewona [7]

Answer:

0.285

Step-by-step explanation:

To do this, divide 28 1/2% by 100%, obtaining:

28.5%

---------- = 0.285

100%

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3 years ago

-1 + 4x=3x +3<br> Solve for x ?

docker41 [41]

Answer:

$x = 4$

Step-by-step explanation:

1. Regroup terms.

$4x - 1 = 3x + 3$

2. Add 1 to both sides.

$4x = 3x + 3 + 1$

3. Simplify 3x + 3 + 1 to 3x + 4.

$4x = 3x + 4$

4. Subtrect 3x from both sides.

$4x - 3x = 4$

5. Simplify 4x - 3x to x.

$x = 4$

Therefor, the answer is, x = 4.

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3 years ago

Need help with this question. Thank you!

Paul [167]

Answer:

61.2

Step-by-step explanation:

Distribute the figure into three shapes. Triangle, and two rectangles.

Find the area.

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3 years ago

Read 2 more answers

Given the function f(x) =-2x+3x^2-5, what is the right end behavior

Leto [7]

The right-end behaviour of all polynomials is always (i.e. whether highest power is odd or even) increasing when the leading coefficient (coefficient of the highest powered variable) is positive.
Otherwise, it is decreasing.

The leading coefficient is +3 (i.e. positive), corresponding to the term +3x^2.

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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:

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

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

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

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

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

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

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

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