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2 years ago

Y + 3y + 5y = -18 what is the value of y

Mathematics

1 answer:

ANEK [815]2 years ago

7 0

Answer:

y = -2

Step-by-step explanation:

Step 1: Combine like terms.

$(y+3y+5y)=-18$
$9y=-18$

Step 2: Divide both sides by 9.

$9y/9=-18/9$
$y=-2$

Step 3: Check if solution is correct.

$-2+3(-2)+5(-2)=-18$
$-2-6-10=-18$
$-18=-18$

Therefore, y = -2.

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What does 2×4+5-3=? Then divide the answer by Five.

Wewaii [24]

Answer:

Step-by-step explanation:

2×4+5-3=

8+5-3=10

10/5=2

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

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

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.

This means that,

Probability that at least one of them needs replacement at the first inspection =

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

5 0

2 years ago

How to solve X^4-19x^2+48=0

ZanzabumX [31]

Two numbers that add up to -19 and multiply to 48 are -16 and -3:

$(x^4-19x^2+48)=0\\(x^2-16)(x^2-3)=0\\(x+4)(x-4)(x^2-3)=0$

So, the solutions come from each parentheses: x+4=0, x-4=0, and x^2-3=0.

x+4=0
x = -4

x-4=0
x = 4

x^2-3=0
x^2 = 3
x = +/- √3

So, the solutions are -4, -√3, √3, and 4.

4 0

3 years ago

gabriel saves 40% of his monthly paycheck for collage. he earned $270 last month.how much money did gabriel save for collage

rusak2 [61]

The best thing to do is to find 10%, and you can do this by dividing $270 by 10, and therefore, $27 is equal to 10%. To find 40%, you've got to multiply $27 by 4, and this gives you $108. Therefore, gabriel saved $108 last month :)
hope this helps u

7 0

3 years ago

If the ratio of boys to girls is 1:4 and there are 20 girls in your class, how many boys are there?

Andre45 [30]

There are 5 boys in the class

4 0

3 years ago

Read 2 more answers