I need help please and thank you

PLEASE HURRY!<br> ------------------------

Levart [38]

Answer:

we cant see the question

4 0

3 years ago

Read 2 more answers

PLEASEEEE HELP ME!!

emmainna [20.7K]

Answer:

A) 5*(-6)*(-1)

-30*(-1)

30

B) -6+(-1)

-6-1

-7

C) 5/(-1)

-5

7 0

4 years ago

6. A submarine has three navigational devices but can remain at sea if at least two are working. Suppose that the failure times

horsena [70]

Answer:

The three survival functions are e-t, e-2/3t and e-1/3t

Then, the rate of the second death equals (multiplying all combinations of 1 survival + 1 death * the rate of death of the third device)

(e-t*(1-e-2/3t)+(1-e-t)e-2/3t)1/3e-1/3t +

(e-t*(1-e-1/3t)+(1-e-t)e-1/3t)2/3e-2/3t +

(e-2/3t*(1-e-1/3t)+(1-e-2/3t)e-1/3t)e-t =

1/3e-4/3t + 1/3e-t -2/3e -2t +

2/3e -5/3t + 2/3e-t -4/3e -2t +

e -4/3 + e -5/3 -2e -2t =

e-t + 4/3e-4/3t + 5/3e-5/3t - 4e-2t

Note that this integrates to 1, thus satisfying one of the rules of a density. We also can see that this function is greater than 0, as e-t + 4/3e-4/3t + 5/3e-5/3t - 4e-2t =

e-t + 4/3e-4/3t + 5/3e-5/3t - e-2t - 4/3e-2t - 5/3e-2t =

e-t - e-2t + 4/3e-4/3t - 4/3e-2t + 5/3e-5/3t- 5/3e-2t =

e-t (1 - e-t) + 4/3e-4/3t(1 - e-2/3t) + 5/3e-5/3t(1 - e-1/3t) > 0 for t > 0

Then, to find the average length that the ship is at sea, we must integrate

t(e-t + 4/3e-4/3t + 5/3e-5/3t - 4e-2t) from 0 to infinity.

However, as a shortcut, note that we can easily identify the integral of the first three functions,

te-t, 4/3te-4/3t, and 5/3te-5/3t, as calculating the expected value of the exponential distribution with parameters 1, 4/3, and 5/3, and the expected value is 1, 3/4, and 3/5

For the final function, 4te-2t = 2*2e-2t , this is twice the expected value of the exponential distribution with parameter 2 = 2 * 1/2 = 1

1 + 3/4 + 3/5 - 1 = 27/20

Step-by-step explanation:

5 0

4 years ago

I need help with this question!

alekssr [168]

Answer:

a) 2(x+12) and 2 times X+ 2 times 6

b) 12y+8

Step-by-step explanation:

for (A) the equation that i choose are the distributed from of the equation

for (B) I combined like terms and they only equation that made sense is this one

5 0

4 years ago

A group of high school athletes has an average GPA of 2.7 with a standard deviation of 0.9. a)Find the percentage of athletes wh

mart [117]

Answer:

a) The percentage of athletes whose GPA more than 1.665 is 87.49%.

b) John's GPA is 3.645.

Step-by-step explanation:

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 2.7, \sigma = 0.9$