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weqwewe [10]
3 years ago
7

The sum of three consecutive even numbers is 78,find the number​

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0

Answer:

24, 26, 28

Step-by-step explanation:

The final number is 78

first we can divide 78/3= 26

so 26+26+26= 78

we can leave the middle 26 but we can give 2 from the first number to the last number

24+26+28=78

(This might be wrong but i tried my best)

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Solve this equation: 8(3x – 6) = 6(4x + 8) A. Solving this equation results in the statement –48 = 48. Because this is a false s
otez555 [7]

Answer:

C. Solving this equation results in the statement –48 = 48. Because this is a false statement, the equation has no solution.

3 0
3 years ago
One online seller has about 6x10to the 8th online orders, and another online seller has about 5x10 to the 7th orders. How many t
Ann [662]

Answer:

100000000

Step-by-step explanation:

6 times 10 to the 8th is 600,000,000

5 times 10 to the 7th is 500,000,000

600,000,000-500,000,000=100,000,000

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3 years ago
Can y’all help plz thx
Alexus [3.1K]

Answer:

C is the answer

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3 years ago
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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
C=100+25h solve for h
aivan3 [116]

Answer:

h=(c/25)-4

Step-by-step explanation:

c-100 = 25h.   "/25"

(c/25)-4 = h

7 0
3 years ago
Read 2 more answers
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