Answer:
The t value for 99% CI for 21 df is 2.831.
The critical value that should be used in constructing the confidence interval is (64.593, 86.407).
Step-by-step explanation:
Now the sample size is less than 30 and also population standard deviation is not known.
Then we will use t distribution to find CI
t value for 99% CI for 21 df is TINV(0.01,21)=2.831
The margin of error is
Hence CI is
Answer: (x) =
Step-by-step explanation:
f(x) = 9x - 8 , to find the inverse of f(x) , replace f(x) with y , then the equation becomes
y = 9x - 8 , then make x , the subject of the formula ,
9x = y + 8
x =
Finally , replace x , with (x) and y with x , so we have
(x) =
Answer:
- 494
Step-by-step explanation:
Breaking down the summation into it;s component parts
= 4k - k(k + 1) ← substitute k = 19
= (4 × 19 ) - 1.5 × (19 × 20)
= 76 - (1.5 × 380)
= 76 - 570
= 494