Answer:
The statements describe transformations performed in f(x) to create g(x) are:
a translation of 5 units up ⇒ c
a vertical stretch with a scale factor of 2 ⇒ d
Step-by-step explanation:
- If f(x) stretched vertically by a scale factor m, then its image g(x) = m·f(x)
- If f(x) translated vertically k units, then its image h(x) = f(x) + k
Let us use these rule to solve the question
∵ f(x) = x²
∵ g(x) is created from f(x) by some transformation
∵ g(x) = 2x² + 5
→ Substitute x² by f(x) in g(x)
∴ g(x) = 2f(x) + 5
→ Compare it with the rules above
∴ m = 2 and k = 5
→ That means f(x) is stretched vertically and translated up
∴ f(x) is stretched vertically by scal factor 2
∴ f(x) is translated 5 uints up
The statements describe transformations performed in f(x) to create g(x) are:
- a translation of 5 units up
- a vertical stretch with a scale factor of 2
Answer:
yes
Step-by-step explanation:
try using a vertical line, if you move the vertical line across the graph and the line touches the graph only at one point, than it means it is a function
Answer:
belongs to the line
. Please see attachment below to know the graph of the line.
Step-by-step explanation:
From Analytical Geometry we know that a line is represented by this formula:

Where:
- Independent variable, dimensionless.
- Dependent variable, dimensionless.
- Slope, dimensionless.
- y-Intercept, dimensionless.
If we know that
,
and
, then we clear slope and solve the resulting expression:



Then, we conclude that point
belongs to the line
, whose graph is presented below.
Answer:
a) ![\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}](https://tex.z-dn.net/?f=%5Cmathrm%7BE%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7B%5Cmathrm%7BH%7D%7D%5E%7B5%7D%20%5Cfrac%7B200%7D%7B101-i%7D)
b) ![\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bk%3D1%7D%5E%7B5%7D%20%5Cfrac%7B%28200%29%5E%7B2%7D%7D%7B%28101-i%29%5E%7B2%7D%7D)
Step-by-step explanation:
Given:
The lifetimes of the individual items are independent exponential random variables.
Mean = 200 hours.
Assume, Ti be the time between (
)st and the
failures. Then, the
are independent with
being exponential with rate
Therefore,
a) ![E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]](https://tex.z-dn.net/?f=E%5BT%5D%3D%5Csum_%7Bi%3D1%7D%5E%7B5%7D%20E%5Cleft%5B%5Ctau_%7Bi%7D%5Cright%5D)

![\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cmathrm%7BE%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7B%5Cmathrm%7BH%7D%7D%5E%7B5%7D%20%5Cfrac%7B200%7D%7B101-i%7D)

The variance is given by, ![\mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bi%3D1%7D%5E%7B5%7D%20%5Cmathrm%7BVar%7D%5BT%5D)
![\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bk%3D1%7D%5E%7B5%7D%20%5Cfrac%7B%28200%29%5E%7B2%7D%7D%7B%28101-i%29%5E%7B2%7D%7D)