Adam tabulated the values for the average speeds on each day of his road trip as 60.5, 63.2, 54.7, 51.6, 72.3, 70.7, 67.2, and 65.4 mph. The sample standard deviation is 7.309.Select the 98% confidence interval for Adam’s set of data. a. 46.94 to 71.33b. 46.94 to 79.46c. 55.45 to 79.46d. 55.45 to 70.95

Option D, 55.45 to 70.95

Step-by-step explanation:

Alpha = 1 – confidence interval  

Alpha = 1 – 0.98 = 0.02

Sample size = n = 8

t (alpha/2) ; (n-1) = t (0.02/2) ; (8-1) = t 0.01, 7 = 2.998

Mean = sum of all frequencies /total number of frequency = 505.6/8 = 63.2

s = 7.309

E = t (0.01;7) * s/sqrt n

Substituting the given values, we get –  

E = 2.998 * 7.309 /sqrt  (8)

E = 7.75  

98% confidence interval  

Mean – E  and Mean + E

63.2 – 7.75 and 63.2 + 7.75  

(55.45, 70.95)

7

step-by-step explanation:

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.