. What is the solution to the equation –3.5(x + 12) =

-3(x+3)^2-3+3x simplify in standard form

IrinaK [193]

Answer: −3x²−15x−30

Step-by-step explanation: use Square of Sum (a+b)²=a²+2ab+b²

−3(x²+2x×3+3²)-3+3x

−(3x² +18x+27)−3+3x

−3x²−18x−27−3+3x

−3x²+(−18x+3x)+(−27−3)

−3x²−15x−30

Hope this helps, now you know the answer and how to do it. Stay healthy and safe and HAVE A BLESSED AND WONDERFUL DAY! :-)

- Cutiepatutie ☺❀❤

6 0

3 years ago

What is the answer to these questions?

Katen [24]

Answer:

1) 8 3/5

2) Option 3

6 0

2 years ago

Read 2 more answers

What is the answer to this 4 1/3 x 6 please help

netineya [11]

Hope this helps...
4 1/3 x 6=
13/3 x 6=
(13 x 6)/3=
13 x 2=
26

6 0

3 years ago

Walt is mixing fruit punch for a party. He combines 1 gallon to quarts 3 pints of orange juice, 1 gallon 3 quarts 7 pints of pin

pochemuha

I think the answer is C because I had da same one in I got it Righ

4 0

3 years ago

The hourly median power (in decibels) of received radio signals transmitted between two cities

trasher [3.6K]

Using the lognormal and the binomial distributions, it is found that:

The 90th percentile of this distribution is of 136 dB.

There is a 0.9147 = 91.47% probability that received power for one of these radio signals is less than 150 decibels.

There is a 0.0065 = 0.65% probability that for 6 of these signals, the received power is less than 150 decibels.

In a lognormal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{\ln{X} - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of $\mu = 3.5$ .

The standard deviation is of $\sigma = \sqrt{1.22}$

Question 1:

The 90th percentile is X when Z has a p-value of 0.9, hence X when Z = 1.28.

$Z = \frac{\ln{X} - \mu}{\sigma}$

$1.28 = \frac{\ln{X} - 3.5}{\sqrt{1.22}}$

$\ln{X} - 3.5 = 1.28\sqrt{1.22}$

$\ln{X} = 1.28\sqrt{1.22} + 3.5$

$e^{\ln{X}} = e^{1.28\sqrt{1.22} + 3.5}$

$X = 136$

The 90th percentile of this distribution is of 136 dB.

Question 2:

The probability is the p-value of Z when X = 150, hence:

$Z = \frac{\ln{X} - \mu}{\sigma}$

$Z = \frac{\ln{150} - 3.5}{\sqrt{1.22}}$

$Z = 1.37$

$Z = 1.37$ has a p-value of 0.9147.

There is a 0.9147 = 91.47% probability that received power for one of these radio signals is less than 150 decibels.

Question 3:

10 signals, hence, the binomial distribution is used.

Binomial probability distribution

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

For this problem, we have that $p = 0.9147, n = 10$ , and we want to find P(X = 6), then:

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

$P(X = 6) = C_{10,6}.(0.9147)^{6}.(0.0853)^{4} = 0.0065$

There is a 0.0065 = 0.65% probability that for 6 of these signals, the received power is less than 150 decibels.

You can learn more about the binomial distribution at brainly.com/question/24863377

5 0

3 years ago

. What is the solution to the equation –3.5(x + 12) = –2x + 18?