Please help me with this! I will give brainlist!

4x - 3(2x + 1) + 7x - 19

Amiraneli [1.4K]

The solution is 5x - 22

8 0

2 years ago

Read 2 more answers

Consider the equation below.

vaieri [72.5K]

Hello there

-y+2= -12(8y+8)

-y+2= -96y+96

-y+96y=96-2

95y=94

y=94/95

:D

6 0

3 years ago

What is the approximate area of a circle with a radius of 30

amid [387]

Answer:

Step-by-step explanation:

Area of circle = π r²

r = 30

30² = 900

π = 3.14 or ≈ 3

A ≈ 900 × 3

A ≈ 2700

so C at 2830 is closest to the answer

8 0

3 years ago

A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January

Blababa [14]

Answer:

a) 27.89% probability that two have ever boycotted goods for ethical reasons

b) 41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) 41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) The expected number is 2.3 and the standard deviation is 1.33.

Step-by-step explanation:

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.

23% of the respondents have boycotted goods for ethical reasons.

This means that $p = 0.23$

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

This is P(X = 2) when n = 6. So

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

$P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789$

27.89% probability that two have ever boycotted goods for ethical reasons

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Either less than two have, or at least two. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$

In which

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

$P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084$

$P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181$

41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

Now n = 10.

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,5}.(0.23)^{5}.(0.77)^{5} = 0.0439$

$P(X = 6) = C_{10,6}.(0.23)^{6}.(0.77)^{4} = 0.0109$

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2343 + 0.1225 + 0.0439 + 0.0109 = 0.4116$

41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

$E(X) = np = 10*0.23 = 2.3$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.23*0.77} = 1.33$

The expected number is 2.3 and the standard deviation is 1.33.

5 0

3 years ago

Review later

____ [38]

It would be 50 times because 15 divided by 3 is 5 and you just add your 0 back and get 50! hope this helps!

3 0

3 years ago