Atire company guarantees that a particular brand of tire has a mean useful lifetime miles or more. a consumer testing agency sampled n = 10 tires on a test wheel that simulated conditions. the sample mean lifetimes (in thousands of miles) were as follows: 42 36 46 43 41 35 43 45 40 39 construct a 98% ci for the mean lifetime of this particular brand of tire.

Step-by-step explanation:

Number of sampled tyres = 10

To determine the mean, we will divide the total number of miles by the total number of tyres.

Mean, u = (42+36+46+43+41+35+43+45+40+39)/10 = 410/10 = 41

Standard deviation = √[summation(u - ub)^2]/n

ub = deviation from the mean, u

Summation(u - ub)^2] = (42-41)^2 + (36-41)^2 + (46-41)^2 +(43-41)^2 + (41-41)^2 + (35-41)^2 + (43-41)^2 + (45 -41)^2 + (40-41)^2 + (39-41)^2

= 1 + 25 + 25 + 4 + 0 + 36 + 4 + 16 + 1 + 4 = 116

Standard deviation = √116/10 = √11.6

= 3.41

We want to determine a 98% confidence interval for the mean mean lifetime of a particular brand of tire.

For a confidence level of 98%, the corresponding z value is 2.33. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

41 +/- 2.33 × 3.41/√10

= 41 +/- 2.33 × 10.78

= 41 +/- 25.11

The lower end of the confidence interval is 41 - 25.11 =15.89

The upper end of the confidence interval is 41 + 25.11 = 66.11

Therefore, with 98% confidence interval, the mean lifetime of a particular brand of tire is between 15.89 miles and 66.11 miles

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

your answer is the frist one

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