Of 375 randomly selected medical students, 30 said that they planned to work in a rural community. find a 95% confidence interval for the true population proportion of all medical students who plan to work in a rural community.

(0.05256, 0.10744) .

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

  (1)

, where = Sample proportion

n= Sample size

z* = Critical z-value.

Let p be the population proportion of all medical students who plan to work in a rural community.

As per given , we have

n= 375

Critical z-value for 95% confidence interval = 1.96

Put all values in (1) , we get

Hence, the 95% confidence interval for the true population proportion of all medical students who plan to work in a rural community is (0.05256, 0.10744) .

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: