Agroup of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. use a 0.05 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. does it appear that students are reasonably good at estimating one minute? 70 79 38 63 44 23 62 61 67 50 61 70 94 87 65

Step-by-step explanation:

Given that a group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below.

Data set is as ollows:

70 79 38 63 44 23 62 61 67 50 61 70 94 87 65

H0: mu = 60 sec

Ha: mu not equals 60 sec

Mean diff = 2.27

Test statistic = 2.27/SE =0.4815

p value =0.6376

Since p>0.05, we accept null hypothesis

i.e. there is statistical evidence to say that students are reasonably good at estimating one minute

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:

cool? what is the actuall ?

step-by-step explanation: