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fgiga [73]
3 years ago
12

3. Graph the line by finding the x and y intercepts. 2x + 3y = 6

Mathematics
1 answer:
lilavasa [31]3 years ago
4 0

Answer:

x-int: (3, 0)

y-int: (0, 2)

Step-by-step explanation:

The x-int is found when y = 0. Set <em>y</em> to 0 to solve for the x-int.

The y-int is found when x = 0. Set <em>x</em> equal to 0 to solve for the y-int.

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Given that f(x)=2x-8, determine the average rate of change over the interval [2, 8].
8_murik_8 [283]

Answer:

2

Step-by-step explanation:

The average rate of change over [2, 8] is the change in the y-values over the change in the x-values.

\frac{f(8) - f(2)}{8 - 2}

f(8) = 2(8) - 8 = 16 - 8 = 8

f(2) = 2(2) - 8 = 4 - 8 = -4

\frac{8 - (-4)}{8-2} = \frac{8 + 4}{8-2} = \frac{12}{6} = 2

5 0
3 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
Find the first four terms of the sequence given by the following an=57-4(n-1),n=1,2,3
Crazy boy [7]

Answer:

57, 53, 49, 45

Step-by-step explanation:

Substitute for the n for n=1,2,3,4 so the first term is 57-4(1-1) so =57-4(0) so

a(1) = 57-0 = 57  so that is the 1st term

2nd term 57-4(2-1) so =57-4(1) so   a(2) = 57-4 = 53, etc.

4 0
3 years ago
Derivatives of inverse trigonometric functions please explain answers
Jobisdone [24]

I'll explain how to do the first one:-

y = cos-1(x2)

This can be described as ' a function of a function'   x^2 is a function of x and cos-1(x^2) is a function of x^2.

We need to  apply the chain rule.

Personally I find this  easier to understand if i let u = x^2, so

If y = f(u) and u is a function of x then

dy/dx = dy/ du * du/dx

Here u = x^2  and y = cos-1(u)

du/dx = 2x

so dy/dx = d(cos-1(x^2) dx = dy/du * du/dx


= -1 / √(1 - u^2) * 2x

= -2x / √(1 - u^2)    

= -2x / √(1 - (x^2)^2)

= -2x / √(1 - x^4)

I hope this helps. but if not. you might like to employ the formulae in the question - The square boxes contain the 'u' s in my answer. These formulae are equivalent to my explanation.

6 0
3 years ago
The graph below represents a population
Nataliya [291]

Answer:

Average rate of  change for the function for the interval (6, 12] is 500 people per year.

Option A is correct.

Step-by-step explanation:

We need to find the average rate of  change for the function for the interval

(6, 12]

The formula used to calculate Average rate of change is:

Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}

We are given a=6 and b=12

Looking at the graph we can see that when x=6 y= 3000 so, f(a)=3000

and when x=12, y=6000 so, f(b)=6000

Putting values in formula and finding Average rate of change:

Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}\\Average \ rate \ of \ change=\frac{6000-3000}{12-6}\\Average \ rate \ of \ change=\frac{3000}{6}\\Average \ rate \ of \ change=500

So, average rate of  change for the function for the interval (6, 12] is 500 people per year.

Option A is correct.

6 0
3 years ago
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