For any three consecutive numbers, prove algebraically that the largest number and the smallest number are factors of the number
1 answer:
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The difference of two squares factoring pattern states that a difference of two squares can be factored as follows:
So, whenever you recognize the two terms of a subtraction to be two squares, you can factor it as the sum of the roots multiplied by the difference of the roots.
In this case, the squares are obvious: is the square of
, and
is the square of
So, we can factor the expression as
(the round parenthesis aren't necessary, I used them only to make clear the two terms)
We can simplify the expression summing like terms:
Answer:
a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.
b) P-value = 0.0055
c) The null hypothesis is rejected.
There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.
d) Critical value tc=-1.96.
As t=-2.55, the null hypothesis is rejected.
Step-by-step explanation:
We have to perform a hypothesis test on the mean.
The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).
a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.
b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.
We can calculate the z-statistic as:
The degrees of freedom are df=599
The P-value for this test statistic is:
c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.
There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.
d) If the significance level is α=0.025, the critical value for the test statistic is t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.
This is the case, as the test statistic is t=-2.55 and falls in the rejection region.
Answer:
10.8
Step-by-step explanation:
We can find the distance using the distance formula:
We then substitute (-3,4) as and (6,-2) as
.