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3 years ago

(3x^3)^2 A. 3x^6 B. 9x^5 C. 6x^6 D. 9x^6 (PLEASE HELPP!!!)

Mathematics

1 answer:

Savatey [412]3 years ago

4 0

Question: Simplify: (3x^3)^2

Step by Step:

STEP 1 : Equation at the end of step 1

(3x³)²

STEP2:

2.1 x³ raised to the 2 nd power = x( 3 * 2 ) = x⁶

Answer: 3²x⁶

Hope it Helps.

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PLEASE HELP MATH!! (BRAINLIEST + MORE) questions in picture!

klasskru [66]

Answer:

A) 2x*2 +6x -3x - 9= 2x*2 +3x -9

B) 2yx*2 +2yx +2y +3x*2 +3x +3

C) 36x*2 y +48yx +3xy*2 + 4y*2

D) 2x*3 - 4x*2 y + 3x -6y

7 0

2 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection = </em>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

How does the graph change between these two functions?

TiliK225 [7]

These are how the graphs will look, they are color coded so have a look at the equations first.

6 0

2 years ago

if a + b + c = 9 and a square + b square + c square = 35 the find the value of ab + bc + ca . I am in need of help

Advocard [28]

a+b+c=9 and a2+b2+c2=35 then by using identity

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca,we get the answer 9x9=35+2(ab+bc+ca)

2(ab+bc+ca)=9x9-35

2(ab+bc+ca) =81-35then we get ab+bc+ca=46/2 so ab+bc+ca=23

8 0

2 years ago

Pls guys help me with this i will give u more points!

gtnhenbr [62]

$3x - 4y = - 6 \\ - 2x + y = 1$

<em>The Ordered pair of solution will be (x,y)=(2,3)</em>

$\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}$

<h3>Steps below⤵️</h3>

Solve the second equation for y
Substitute the given value of y into the first equation
Solve the first equation for x
Substitute the given value of x into the second equation
Solve the equation for y
The possible solution of the system is the ordered pair (x,y)
And we are done solving!!~

5 0

2 years ago