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Dennis_Churaev [7]

3 years ago

8

there are 24 apple trees in my orchard. if i can harvest the fruit from 2 and 2 thirds trees each day, how many days will it tak

e to harvest the entire orchard?

1 answer:

valentina_108 [34]3 years ago

8 0

Answer: 9 days.

Step-by-step explanation:

2 2/3 is the amount able to be harvested, you want to find days. So, 2 2/3(d)=9. You need to solve for days (d) which equals 9.

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What is the probability of picking an orange marble and flipping tails? 1/7 1/14 2/14 8/14

vladimir2022 [97]

Answer:

Flipping tails would be 1/2 but how many marbles are there and how many orange ones are there?

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the missing side of this right

navik [9.2K]

Answer:

x = 13

Step-by-step explanation:

First start with the blank equation

$a^{2} + b^{2} = c^{2}$

Plug in the legs and hypotenuse. Remember c squared is always the hypotenuse.

$5^{2} + 12^{2} = c^{2}$

Then solve

25 + 144 = $c^{2}$

169 = $c^{2}$

$\sqrt{169} = \sqrt{c^{2} }$

c = 13

3 0

3 years ago

Does anyone know how to do #4 plz help me!!!

liberstina [14]

If you can get me some numbers this would be easier
however there's r= radius
h=height
I'm not fully understanding this

4 0

3 years ago

Read 2 more answers

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

I need help asap :):)):

slega [8]

Answer:

Its to multiply the numerators by each other and the denominators by each other

Step-by-step explanation:

If adding and subtracting fractions is about finding the least common denominator

dividing fractions is to keep the first fraction,multiply and reverse the second fraction,so dividing is pretty similar to multiplying but just switching the divide sign with a multiplication sign and reversing the second fraction from numerator/denominator to denominator/numerator

and there is always a big difference if you multiply/divide or add/subtract

4 0

3 years ago