Help please if you can

There are 19 boxes of cherries. This is 5 less than twice the number of boxes of kiwis. How many boxes of kiwis are there?

sasho [114]

All you have to do is 19+5 because it is 5 less. 19 is 5 less than the kiwis. so 19+5= 24.

There is 24 boxes of kiwis

8 0

3 years ago

Read 2 more answers

Suppose 12% of the students chose to study French their freshman year, and that meant that there were 21 such students. How many

AleksAgata [21]

Answer:

154 students

Step-by-step explanation:

First get the total number of students .

This can be gotten by

12% of A = 21

Where A represents the total number of students.

12% represents the % of A that chose to study French and 21 is the number of students that studied French .

Therefore,

12% /100% x A = 21

0.12 x A = 21

Divide both sides by 0.12

0.12/0.12 x A = 21/0.12

A = 175

The total number of students is 175.

If 21 chose to study French their freshman year ,number of students that chose not to will be total number of students minus number of those who chose to study French.

That’s

175 - 21

= 154

154 students chose not to study French their freshman year

4 0

3 years ago

What is 100x82727=x?

Nadusha1986 [10]

Answer:

8272700

Step-by-step explanation:

Just add two zeros in the end

4 0

2 years ago

Read 2 more answers

(3x + 1) > -7+ 3x Plz help me

Luda [366]

Answer:

0, -1

Step-by-step explanation:

This question was answered in another place since you asked it twice. Please refer to the other place that you asked the same question.

5 0

2 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

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

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.

This means that,

Probability that at least one of them needs replacement at the first inspection =

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

5 0

3 years ago