Write the expression for when you translate the graph of y = |3x− 5| one unit to the right. Plzz help

Mathematics

2 answers:

AfilCa [17]3 years ago

6 0

Answer:

2<x<4/3

Step-by-step explanation:

Given the equation of a graph to be y = |3x− 5|, if the equation is one unit to the right, this can be expressed as |3x-5| > 1.

Solving the resulting equation

|3x-5| > 1.

Since the function 3x-5 is in a modulus sign, this means that the function can take both negative and positive values.

For positive value of the function;

+(3x-5) > 1

3x > 1+5

3x>6

x>6/3

x>2 ... (1)

For the negative value of the function;

-(3x-5) > 1

On expansion

-3x+5 > 1

-3x > 1-5

-3x > -4

Multiplying through by -1 will also change the inequality sign

x < -4/-3

x < 4/3...(2)

Combining equation 1 and 2, we have;

2<x<4/3

12345 [234]3 years ago

3 0

Answer:

y=|3x-6|

Step-by-step explanation:

on unit to the right is -1 inside the absolute value bracket so it would be y=|3x-5-1| which simplifies to y=|3x-6|

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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.