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3 years ago

NEED HELP ASAP BRAINLIEST ANSWER IF CORRECT

1 answer:

4 0

If we need sections that are 18 ft. long from available 78 ft. long line than we will divide 78/18 =4.(3)
So we will get 4 sections that are 18 ft. long and we will have a remainder of 6 ft line not used.

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Which linear function represents the table? A) y = 2x − 5 B) y = 2x + 5 C) y = −2x + 5 D) y = −2x − 5

lubasha [3.4K]

Answer:

y= -2x - 1

Step-by-step explanation:

that is the answer sorry if wrong

5 0

3 years ago

2. Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,00

svlad2 [7]

Using the binomial distribution, it is found that:

a) There is a 0.0501 = 5.01% probability that you need to contact four people.

b) You expect to contact 1.82 students until you find one who lives within five miles of you.

c) The standard deviation is of 1.22 students.

d) There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

e) It is expected that 2.75 students live within five miles of you.

For each student, there are only two possible outcomes. Either they live within 5 miles of you, or they do not. The probability of a student living within 5 miles of you is independent of any other student, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

55% of the students live within five miles of you, thus $p = 0.55$ .

Item a:

This probability is P(X = 0) when n = 3(none of the first three living within five miles of you) multiplied by 0.55(the fourth does live within five miles), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125$

$p = 0.091125(0.55) = 0.0501$

0.0501 = 5.01% probability that you need to contact four people.

Item b:

The expected number of trials in the binomial distribution until q successes is given by:

$E = \frac{q}{p}$

In this problem, $p = 0.55$ , and 1 trial, thus $q = 1$ , hence:

$E = \frac{1}{0.55} = 1.82$

You expect to contact 1.82 students until you find one who lives within five miles of you.

Item c:

The standard deviation of the number of trials until q successes are found is given by:

$S = \frac{\sqrt{q(1 - p)}}{p}$

Hence, since $q = 1, p = 0.55$ :

$S = \frac{\sqrt{0.45}}{0.55} = 1.22$

The standard deviation is of 1.22 students.

Item d:

This probability is $P(X = 3)$ when $n = 5$ , hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.55)^{3}.(0.45)^{2} = 0.3369$

There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

Item e:

The expected value of the binomial distribution is:

$E(X) = np$

Hence, since $n = 5, p = 0.55$ :

$E(X) = 5(0.55) = 2,75$

It is expected that 2.75 students live within five miles of you.

A similar problem is given at brainly.com/question/25343741

7 0

3 years ago

The sale price of a swimming pool after 16.5% discount $1135.60. Find the original price of the pool

OverLord2011 [107]

well, let's say the original price is "x", and thus is also the 100% price of the item, however the item is discounted by 16.5%, well 100% - 16.5% = 83.5%, so then the discounted price is really the 83.5% of the item, and we know that is $1135.60.

now, if 1135.60 is 83.5%, what is the 100% or namely "x"?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 1135.60&83.5\\ x&100 \end{array}\implies \cfrac{1135.60}{x}=\cfrac{83.5}{100}\implies 113560 = 83.5x \\\\\\ \cfrac{113560}{83.5}=x\implies 1360 = x$