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3 years ago

Rewrite the formula for the circumference of a circle C=2xpix r in terms of its diameter d. Show your work.

2 answers:

ozzi3 years ago

7 0

Hi there!

The answer is
$circumference \: = d\pi$
In this formula d represents the diameter.

We know that 2 times the radius equals the diameter, hence
$2r = d$

Therefore we can replace the 2r in the original formula by d, and then we get the following:
$circumference = 2\pi \: r = 2r\pi = d\pi$

Keith_Richards [23]3 years ago

3 0

Diameter = 2·radius

С = 2·π·r ⇒ С = π·d

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Ana's class earns points for helping others. The graph shows the number of points the class earned each day for a week.How many

algol [13]

Answer:Gimme a sec

Step-by-step explanation:

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3 years ago

Which is the BEST estimate for the value of x?

Lapatulllka [165]

To solve this, I would square root all of the possible answers and see which ones the arrow falls between. After that I would choose the answer closest to it.

Hope this helps!

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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>

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

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3 years ago

Simplify the expression 8 + (7 + x)

baherus [9]

Answer:

15 + x

Step-by-step explanation:

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2 years ago

I need help with this please

balandron [24]

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3 years ago