If the graph of f(x) = x^2, how will the graph be affected if it is changed to f(x) = 3r^2?

2 answers:

7 0

Answer:

It vertically stretches the graph.

Explanation:

If you multiply the function by a constant , a ,
if a > 1 the graph will be stretched
If 0If a<0 there will be a combination of vertical stretch or compression with vertical reflection

Eva8 [605]3 years ago

3 0

Answer:

the graph curve will goes above f(x)=x^2.f(x)=3*x^2 curve will also give higher value same value of x .

Step-by-step explanation:

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Read 2 more answers

Suppose that the probability that any particle emitted by a radioactive material will penetrate a certain shield is 0.01. If 10

AnnyKZ [126]

Answer:

a)P=0.42

b) $n\geq 297$

Step-by-step explanation:

We have a binomial distribution, since the result of each experiment admits only two categories (success and failure) and the value of both possibilities is constant in all experiments. The probability of getting k successes in n trials is given by:

$P=\begin{pmatrix}n\\ k\end{pmatrix} p^k(1-p)^{n-k}=\frac{n!}{k!(n!-k!)}p^k(1-p)^{n-k}$

a) we have k=2, n=10 and p=0.01:

$P=\frac{10!}{2!(10!-2!)}0.01^2(1-0.01)^{10-2}\\P=\frac{10!}{2!*8!}0.01^2(0.99)^{8}\\P=45*0.01^2(0.99)^8=0.42$

b) We have, $1-(1-p)^n=P$ , Here P is the probability that at least one particle will penetrate the shield, this probabity has to be equal or greater than 0.95. Therefore, this will be equal to subtract from the total probability, the probability that the particles do not penetrate raised to the total number of particles.