Answer:
yes
Step-by-step explanation:
Answer:
Step-by-step explanation:
The hypothesis is written as follows
For the null hypothesis,
µd ≤ 10
For the alternative hypothesis,
µ > 10
This is a right tailed test
Since no population standard deviation is given, the distribution is a student's t.
Since n = 97
Degrees of freedom, df = n - 1 = 97 - 1 = 96
t = (x - µ)/(s/√n)
Where
x = sample mean = 8.9
µ = population mean = 10
s = samples standard deviation = 3.6
t = (8.9 - 10)/(3.6/√97) = - 3
We would determine the p value using the t test calculator. It becomes
p = 0.00172
Since alpha, 0.01 > than the p value, 0.00172, then we would reject the null hypothesis. Therefore, At a 1% level of significance, there is enough evidence that the data do not support the vendor’s claim.
Answer:
and 
.
Step-by-step explanation:
So I believe the problem is this:
where we are asked to find values for 
and 
such that the equation holds for any 
in the equation's domain.
So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).
In other words this will clear the fractions.
As you can see there was some cancellation.
I'm going to plug in -7 for x because x+7 becomes 0 then.
Divide both sides by -10:
Now we have:
with
I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.
Divide both sides by 10:
So and .
Answer:
If you want to spend no less than $40, and you want to spend twice as much on the first gift (gift x), you can spend $26 on gift x. On the second gift (gift y), you can spend $13. You will spend a total of $39 on both gifts which is less than $40 and have spend twice as much on the first gift as you did the second ($13 x 2= $26).
