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2 years ago

Business Loss A company loses $144 as a result of a shipping delay. The 9 owners of the

Mathematics

1 answer:

koban [17]2 years ago

5 0

Answer:

Each person will earn $16.

Step-by-step explanation:

Given that:

Amount of loss faced by the company = $144

Number of people who would share the amount = 9

Let,

x be the share of each person.

Number of people * Share per person = Total loss

9x = 144

Dividing both sides by 9

$\frac{9x}{9}=\frac{144}{9}\\$

x = $16

Hence,

Each person will earn $16.

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Please answer number 1 and 2.

zavuch27 [327]

1, a.) The two specific conjectures are in the first image.

1, b.) The two general conjectures are in the second image.

2, a.) Two specific conjectures for this pattern are:

The common difference between two consecutive terms is 3.
And the given difference is A.P.

2, b.) From our observation two general conjecture is that the given sequence is an arithmetic sequence and the common difference is 3.

For finding its nth term we can use the formula: a(n) = a + (n-1) x d.

2, c.) A formula for the given pattern is 5 + (n-1)3, where n is the number of the term.

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2 years ago

To divide a decimal number by 1,000, move the decimal point _____.

valkas [14]

The answer is 3 places to the left

6 0

3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago

A recipe for fruit punch calls for every 9 cups of grapefruit juice for every 4 cups of OJ. If 60 cups of OJ are used, then how

vampirchik [111]

Answer:

15 cups

Step-by-step explanation:

60/4=15

3 0

2 years ago

140 subtracted from the square of a number the result is three times the number find the negative solution

natima [27]

Answer:

$Suppose~that~the~number~is~x.\\According~to~question,\\x^2-140=3x\\or, x^2-3x-140=0\\From~the~quadratic~formula,\\x = \frac{-(-3)+\sqrt{(-3)^2-4(1)(-140)} }{2(1)} ~and ~x= \frac{-(-3)-\sqrt{(-3)^2-4(1)(-140)} }{2(1)} \\or, x = \frac{3+\sqrt{9+560} }{2}~and~x=\frac{3-\sqrt{9+569} }{2}\\or, x = \frac{3+\sqrt{569} }{2}~and~x=\frac{3-\sqrt{569} }{2}\\So,~the~required~negative~number~is~\frac{3-\sqrt{569} }{2}.$

5 0

2 years ago