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2 years ago

Which equation is the slope intercept form of this equation?

Mathematics

1 answer:

AnnZ [28]2 years ago

7 0

Answer:

Step-by-step explanation:

Y - 4 = -2x + 10 Multiply -2 to the inside the brackets

Y = -2x + 14 Add the 4 over

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Robin and Shaon got 5/6 and 4/5 of the total marks. Robin obtained 30 more marks than Shaon.What is the total marks?

Natasha2012 [34]

Answer:

Robin= 750

Sharon= 720

Step-by-step explanation:

let the total marks obtained be X

from question, Robin = 5X/6

Shaon= 4X/5

then, 5X/6= 4X/5+30

or, 5X/6= 4X+150/5

or 25X=24x+900

or,X= 900

therefore robin=750

Shaon = 720

3 0

2 years ago

Choose the expression that is equivalent to both −x/y and x/−y.

astraxan [27]

Answer:

-x/-y

Step-by-step explanation:

3 0

3 years ago

F(x)+x^2-3x-2x is shifted 4 units left. The result is g(x). What is g(x)

otez555 [7]

Answer:

g(x) = x² + 5x + 2

Step-by-step explanation:

If a function is f(x) = x²

And this function is shifted 4 units left then the rule to be followed is,

g(x) = f(x + 4)

Following the same rule,

For a function f(x) = x² - 3x - 2 when shifted 4 units to the left,

g(x) = f(x + 4)

g(x) = (x + 4)² - 3(x + 4) - 2

= x² + 8x + 16 - 3x - 12 - 2

= x² + 5x + 2

Therefore, g(x) = x² + 5x + 2 will be the transformed function.

3 0

3 years ago

A red apple has mass of 20 grams .you cut it into 5 slices. what about the aplles has changed ?

Nutka1998 [239]

Well you have cut it into smaller pieces so it would be I guess 4 grams now because 20 divided by 5 is 4.

7 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