College students and STDs: A recent report estimated that 25% of all college students in the United States have a sexually transmitted disease (STD). Due to the demographics of the community, the director of the campus health center believes that the proportion of students who have a STD is higher at his college. He tests H 0: p = 0.25 versus H a: p > 0.25. The campus health center staff select a random sample of 150 students and determine that 43 have been diagnosed with a STD. Conduct a hypothesis test to address the director’s hypothesis. Use a 5% significance level to make your decision. Use the applet (at the top of this Checkpoint) to determine the P‐value. Which of the following is an appropriate conclusion based on the results? Group of answer choices The data provides strong evidence that the proportion of students at his college who have a STD is more than 25%. The data suggests that the proportion of students at his college who have a STD is 29%. Of the students surveyed at his college, 29% have an STD, but this is not strong enough evidence to conclude that the proportion of students at his college who have a STD is higher than the report’s estimate. This survey suggests that 25% of students at his college have a STD.

i need more info. are you looked to see there mph? how much they deliver? how far they go?

step-by-step explanation:

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %

answer: the answer is 110

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)