You plug in f(x) into where x is within g(x) so you get G(f(x))=(X+6)^4
Of 85% of cars sold where not black then 100-85% where black = 15%
let the number of cars sold be y
then 15%*y=300
y= 300 divided by 15%
= (300*100)/15
=2000
therefore 2000 cars were sold
(Простите, пожалуйста, мой английский. Русский не мой родной язык. Надеюсь, у вас есть способ перевести это решение. Если нет, возможно, прилагаемое изображение объяснит достаточно.)
Use the shell method. Each shell has a height of 3 - 3/4 <em>y</em> ², radius <em>y</em>, and thickness ∆<em>y</em>, thus contributing an area of 2<em>π</em> <em>y</em> (3 - 3/4 <em>y</em> ²). The total volume of the solid is going to be the sum of infinitely many such shells with 0 ≤ <em>y</em> ≤ 2, thus given by the integral

Or use the disk method. (In the attachment, assume the height is very small.) Each disk has a radius of √(4/3 <em>x</em>), thus contributing an area of <em>π</em> (√(4/3 <em>x</em>))² = 4<em>π</em>/3 <em>x</em>. The total volume of the solid is the sum of infinitely many such disks with 0 ≤ <em>x</em> ≤ 3, or by the integral

Using either method, the volume is 6<em>π</em> ≈ 18,85. I do not know why your textbook gives a solution of 90,43. Perhaps I've misunderstood what it is you're supposed to calculate? On the other hand, textbooks are known to have typographical errors from time to time...
Answer:
CI 95%(μ)= [13.506 ; 14.094]
Step-by-step explanation:
The confidence interval (CI) formula is:
CI (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]
alpha= is the proportion of the distribution tails that are outside the confidence interval. In this case, 5% because 100-90%
Z(5%/2)= is the critical value of the standardized normal distribution. In this case is 1.96
σ= standard deviation. In this case 0.75 day
mean= 13.8 days
n= number of observations
. In this case 25
Then, the confidence interval (90%) is:
CI 95%(μ)= 13.8+- [1.96*(0.75/sqrt(25)]
CI 95%(μ)= 13.8+- [1.96*(0.75/5)
]
CI 95%(μ)= 13.8+- (0.294)
CI 95%(μ)= [13.8-0.294 ; 13.8+0.294]
CI 95%(μ)= [13.506 ; 14.094]
A change to the inside of the function notation is always a left/right or horizontal change to the graph.
Answer: A
Subtracting 5 actually shifts the graph to the right by 5 units. It's counter-intuitive and there are many ways to explain why it's the opposite of what you think it'd do, but it is a shift to the right by 5 units.
PS
A change outside the f(x) notation is an up/down or vertical change to the graph.