Help me with number 3 pleaseee

The graph of y=x^3 is transformed as shown in the graph below. Which equation represents the transformed function?

S_A_V [24]

Y=(-x)^3-4 the last one hope this helped on time!! If u ever need help with this u can use the desomos calculator it’s perfect for graphing good luck!

4 0

1 year 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

2 years ago

Kelley Katherine takes a trip to Paris. When she was standing near the 1063 foot tall Eiffel tower, she noticed that her shadow

alina1380 [7]

In order to complete this, we must write it as a ratio. This would look like: 2.4:5.25=x:1063. In order to solve for x, we must first divide 5.25 by 2.4, which I will mark as y, then we take y and divide it into 1063, giving us the answer of 485.94, rounded will give you 486. x now equals 486.

5 0

3 years ago

Bonnie is making a dipping sauce. She mixes 150 milliliters of soy sauce with 100 milliliters of vinegar. How much soy sauce doe

oksian1 [2.3K]

Answer:

Bonnie is mixing 1.50 milliliter of soy sauce with 1 milliliter of vinegar

Step-by-step explanation:

Given

Bonnie mixes 150 milliliters of soy sauce with 100 milliliters of vinegar

So, in order to find the amount of soy sauce mixed with one ml of vinegar, we have to divide the amount of soy sauce by vinegar.

So,

$Soy\ sauce\ mixed\ with\ 1\ ml\ of\ vinegar\ =\frac{150}{100}\\ = \frac{3}{2}\\=1.50$

Hence,

Bonnie is mixing 1.50 milliliter of soy sauce with 1 milliliter of vinegar

3 0

3 years ago

A potato sorting machine produces 5 pounds of potatoes an average for the standard aviation is a half a pound assuming the weigh

mixas84 [53]

For a 95% confidence interval, the corresponding z-score is 1.96. Therefore the deviation will by 1.96*0.5 lbs = 0.98 lbs. Therefore, the confidence interval will be (5 - 0.98, 5 + 0.98), which is (4.02, 5.98). The weight range is from 4.02 lbs to 5.98 lbs.

5 0

3 years ago