An advocacy group claims that the mean braking distance of a certain type of tire is 75 feet when the car is going 40 miles per hour. In a test of 80 of these tires, the braking distance has a mean of 77 and a standard deviation of 5.9 feet. Find the standardized test statistic and the corresponding p-value.

The calculated value t = 3.032 >  1.9904 at 5% level of significance

Step-by-step explanation:

Step(i):-

Given mean of the population 'μ'=75 feet

Given mean of the sample(x⁻) = 77

Given standard deviation of the sample(S) = 5.9

Given sample size 'n' =80

Step(ii):-

Test statistic

            t = 3.032

Degrees of freedom ν=n-1 = 80-1 =79

tabulated value = 1.9904 at 5% level of significance

The calculated value t = 3.032

Final -

The calculated value t = 3.032 >  1.9904 at 5% level of significance

false

step-by-step explanation:

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺