Aparticular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. from past studies of this tire, the standard deviation is known to be 8,000. a survey of owners of that tire design is conducted. of the 31 tires surveyed, the mean lifespan was 46,800 miles with a standard deviation of 9,800 miles. using alpha do the data support the claim at the 5% level?

Let be the distance traveled by deluxe tire .

As per given , we have

Null hypothesis :

Alternative hypothesis :

Since is left-tailed and population standard deviation is known, thus we should perform left-tailed z-test.

Test statistic :

where, n= sample size

= sample mean

= Population mean

=sample standard deviation

For , we have

By using z-value table,

P-value for left tailed test : P(z≤-2.23)=1-P(z<2.23)   [∵P(Z≤-z)=1-P(Z≤z)]

=1-0.9871=0.0129

 Decision : Since p value (0.0129) < significance level  (0.05), so we reject the null hypothesis .

[We reject the null hypothesis when p-value is less than the significance level .]

Conclusion : We do not have enough evidence at 0.05 significance level to support the claim that t its deluxe tire averages at least 50,000 miles before it needs to be replaced.

answer: y=4000x+14000

step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)