Answer:
Step-by-step explanation:
Given
+ 3(
- 
)²
= + 3 ( )²
= + 3 × 
= +
=
Not sure sorry I will look at it again though
Answer:
2a^4+5a^3-6a^2+19a-20
Step-by-step explanation:
(a^2+3a-4)(2a^2-a+5)
2a^4-a^3+5a^2+6a^3-3a^2+15a-8a^2+4a-20
2a^4-a^3+6a^3+5a^2-3a^2-8a^2+15a+4a-20
2a^4+5a^3+2a^2-8a^2+19a-20
2a^4+5a^3-6a^2+19a-20
For large sample confidence intervals about the mean you have:
xBar ± z * sx / sqrt(n)
where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.
z * sx / sqrt(n) = width.
so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348
The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758
Since n must be integer valued we need to take the ceiling of this solution.
n = 22
The triangle ADC and the triangle CDB are similar. Therefore:
Answer: A 4.5
