Solve the quadratic equation numerically - 4x = -x + 12
1 answer:
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Recursive Formula
The top row says the first term is 8
The bottom row says that to get the nth term, we subtract 7 from the (n-1)th term. So basically we subtract 7 from each term to get the next term.
Note the subscripts tell us which term we're working with.
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Arithmetic Formula
We have a = 8 as the first term and d = -7 as the common difference.
a(n) = a + d(n-1)
a(n) = 8 + (-7)(n-1)
a(n) = 8 - 7n + 7
a(n) = -7n+15
The nth term arithmetic formula is a(n) = -7n+15
If you plug in n = 1, you should get a(n) = 8
If you plug in n = 2, you should get a(n) = 1
and so on.
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Finding the 10th term
Plug in n = 10 to get
a(n) = -7n+15
a(10) = -7(10)+15
a(10) = -70+15
a(10) = -55
The 10th term is -55
Answer:
Since is a right tailed test we need to look on the t distribution with 12 degrees of freedom that accumulates 0.05 of the area on the right. And we can use the following excel code:
"=T.INV(0.95,12)" and we got
Step-by-step explanation:
When we have two independent samples from two normal distributions with equal variances we are assuming that
And the statistic is given by this formula:
Where t follows a t distribution with degrees of freedom and the pooled variance
is given by this formula:
This last one is an unbiased estimator of the common variance
The system of hypothesis on this case are:
Null hypothesis:
Alternative hypothesis:
Or equivalently:
Null hypothesis:
Alternative hypothesis:
Our notation on this case :
represent the sample size for group Edne
represent the sample size for group Orno
represent the sample mean for the group Edne
represent the sample mean for the group Orno
represent the sample standard deviation for group Edne
represent the sample standard deviation for group Orno
First we can begin finding the pooled variance:
And the deviation would be just the square root of the variance:
And now we can calculate the statistic:
Now we can calculate the degrees of freedom given by:
Since is a right tailed test we need to look on the t distribution with 12 degrees of freedom that accumulates 0.05 of the area on the right. And we can use the following excel code:
"=T.INV(0.95,12)" and we got
Since our calculated value is higher than the critical value we have anough evidence to reject the null hypothesis, and on thsi case the mean for Orno is significanlty higher than the mean for Edne at 5% of significance.