Can’t figure this out help please

Errors in an experimental transmission channel are found when thetransmission is checked by a certifier that detects missing pul

Ivahew [28]

Answer:

a) $P(X \leq 4)$

And we can find this using the cumulative distribution function:

$P(X \leq 4) = F(4) = 0.9$

b) $P(X > 7)$

And we can find this using the cumulative distribution function and the complement rule on this way:

$P(X >7) =1-P(X\leq 7)= 1- F(7) = 1-1 = 0$

c) $P(X \leq 5)$

And we can find this using the cumulative distribution function:

$P(X \leq 5) = F(5) = 0.9$

d) $P(X > 4)$

And we can find this using the cumulative distribution function and the complement rule on this way:

$P(X >4) =1-P(X\leq 4)= 1- F(4) = 1-0.9 = 0.1$

e) $P(X \leq 2)$

And we can find this using the cumulative distribution function:

$P(X \leq 2) = F(2) = 0.7$

Step-by-step explanation:

For this case we have the following cumulative distribution function:

$F(x) = 0 , x$

$F(x) = 0.7, 1 \leq x$

$F(x) = 0.9, 4 \leq x$

$F(x) = 1, x \geq 7$

Part a

We want this probability:

$P(X \leq 4)$

And we can find this using the cumulative distribution function:

$P(X \leq 4) = F(4) = 0.9$

Part b

We want this probability:

$P(X > 7)$

And we can find this using the cumulative distribution function and the complement rule on this way:

$P(X >7) =1-P(X\leq 7)= 1- F(7) = 1-1 = 0$

Part c

We want this probability:

$P(X \leq 5)$

And we can find this using the cumulative distribution function:

$P(X \leq 5) = F(5) = 0.9$

Part d

We want this probability:

$P(X > 4)$

And we can find this using the cumulative distribution function and the complement rule on this way:

$P(X >4) =1-P(X\leq 4)= 1- F(4) = 1-0.9 = 0.1$

Part e

We want this probability:

$P(X \leq 2)$

And we can find this using the cumulative distribution function:

$P(X \leq 2) = F(2) = 0.7$

7 0

3 years ago

Which statement is an example of the reflexive property of congruence?

marusya05 [52]

C. EFG = EFG

The reflexive property allows you to change the order as long as they are the same.

3 0

4 years ago

Read 2 more answers

Finn changes his mind and, from now on, decides to take the normal route to work everyday. On any given day, the time (in minute

Margaret [11]

Answer:

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

Step-by-step explanation:

This can be solved by the the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

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

Each z-score value has an equivalent p-value, that represents the percentile that the value X is:

The problem states that:

Mean = 35, so $\mu = 35$

Variance = 81. The standard deviation is the square root of the variance, so $\sigma = \sqrt{81} = 9$ .

Find the 33rd percentile of the time it takes Finn to get to work on any given day. Do not include any units in your answer.

Looking at the z-score table, $z = -0.44$ has a pvalue of 0.333. So what is the value of X when $z = -0.44$ .

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

$-0.44 = \frac{X - 35}{9}$

$X - 35 = -3.96$

$X = 31.04$

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

Over the next 2 days, find the probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that Finn took more than 40.5 minutes to get to work on the first day. The first step to solve this problem is finding the z-value of $X = 40.5$ .

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

$Z = \frac{40.5 - 35}{9}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.7291. This means that the probability that it took LESS than 40.5 minutes for Finn to get to work is 72.91%. The probability that it took more than 40.5 minutes if $P_{1} = 100% - 72.91% = 27.09% = 0.2709$

$P_{2}$ is the probability that Finn took more than 38.5 minutes to get to work on the second day. Sine the probabilities are independent, we can solve it the same way we did for the first day, we find the z-score of

$X = 38.5$

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

$Z = \frac{38.5 - 35}{9}$

$Z = 0.39$

$Z = 0.39$ has a pvalue of 0.6517. This means that the probability that it took LESS than 38.5 minutes for Finn to get to work is 65.17%. The probability that it took more than 38 minutes if $P_{1} = 100% - 65.17% = 34.83% = 0.3483$

So:

$P = P_{1} + P_{2} = 0.2709 + 0.3483 = 0.6192$

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

7 0

4 years ago

WILL MARK BRAINLIEST

seropon [69]

C. (22,16)

Y=x-6
16=22-6
16=16

7 0

3 years ago

supposed f(x)=2x-6 . write the equation for g(x) if g(c)=f(x)+3 . how did the transformation shift the graph of f(c) ?

umka21 [38]

Answer:

g(c)=2x-3

Step-by-step explanation:

given f(x)=2x-6 put the value of f(x) in g(x)

g(x)=f(x)+3=2x-6+3

g(x)=2x-3

5 0

3 years ago