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3 years ago

The diameter of a circle is 10 units. What is the radius of the circle? units

Mathematics

2 answers:

tia_tia [17]3 years ago

7 0

Answer:

The answer is 5.

Step-by-step explanation:

Hope this helped Mark BRAINLEST!!!

egoroff_w [7]3 years ago

7 0

Answer:

Step-by-step explanation:

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Louisa needs 3 liters of lemonade and punch for a picnic she has 1800 milliliters of lemonade. How much punch does she need expl

8090 [49]

There are 1000 millimetres in a litre, so 1800 ml is 1.8 litres. 3-1.8 is 1.2.

5 0

3 years ago

Read 2 more answers

Simplify (5x^3 - 7x^2 + 3x - 4) - (8x^3 + 2x^2 + 3x - 7)

muminat

Answer:

-3x^4 - 14x^3 + 3x^2 - 10

Step-by-step explanation:

6 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>

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