Richard is planning a banquet for the wrestling

Help me with my math and please show your work

LenaWriter [7]

Answer:

The solution is:

$160 for booth + $16 needles = $176 as capital

$15 price to sell each scarf - $4 cost of the yarn to make

the scarf = $11 as profit

$176 / $11 = 16

The students must sell 16 pieces of scarves to get the

break even point of $176 for the capital

6 0

3 years ago

Name 5 decimal numbers that are between 4.78 and 4.79. Explain your answer.

Karolina [17]

4.781, 4.782,4.783,4.784,4.785

I can't explain though:(

6 0

3 years ago

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A smaller triangle is cut out of a larger triangle. The larger triangle is a right triangle with side lengths of 5 millimeters a

iren [92.7K]

Answer:

Sorry if this is late, but it's 24mm2

Step-by-step explanation:

i got a 100 on the quiz

8 0

3 years ago

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3. Which expression is equivalent to log532?

saul85 [17]

Answer: The Nth power xN of a number x was originally defined as x multiplied by itself, until there is a total of N identical factors. By means of various generalizations, the definition can be extended for any value of N that is any real number.

(2) The logarithm (to base 10) of any number x is defined as the power N such that

x = 10N

(3) Properties of logarithms:

(a) The logarithm of a product P.Q is the sum of the logarithms of the factors

log (PQ) = log P + log Q

(b) The logarithm of a quotient P / Q is the difference of the logarithms of the factors

log (P / Q) = log P – log Q

(c) The logarithm of a number P raised to power Q is Q.logP

log[PQ] = Q.logP

Step-by-step explanation:

8 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