Answer:
dx. = yex ⇒ dy y. ∫. = x ... 2 xx dx. ∫. , i.e. yx = 4. 4 x c. +. Hence y = 3. 4 x c x.
Answer:
kinda late but the answer is B
Hope this helps!! <3
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
Answer: 59.92
Step-by-step explanation:
90 + 50 = 140
140 - 40% = 56
56 + 7% = 59.92
f(-9) means x is -9
Replace x with -9 and solve:
2/3(-9) - 6
Simplify:
-6 - 6 = -12
f(-9) = -12
