Aschool district claims that the average teacher in the district earns $45,000 per year. the teacher's union disputes this claim and argues that the average salary is actually less. a random sample of 20 teachers yields a mean salary of $44,500 with a sample standard deviation of $1,750. what's the p­value for a test of the hypothesis that h0 : m = 44,5 00 and ha : m < 44,500?

a. 01 < p < .02
b. 02 < p < .025
c. 025 < p < .05
d. 05 < p < .10
e. 10 < p < .15

Option e) 0.10 < P < 0.15

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $45,000

Sample mean, = $44,500

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = $1,750

First, we design the null and the alternate hypothesis

We use one-tailed(left) t test to perform this hypothesis.

Formula:

Putting all the values, we have

Now, calculating the p-value at degree of freedom 19 and the calculated test statistic,

p-value = 0.108494

Thus,

Option e) 0.10 < P < 0.15

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)