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Elena-2011 [213]
3 years ago
6

HELP!!!!! What is the exclusion of 57x^5/19x^2??

Mathematics
1 answer:
Setler [38]3 years ago
3 0
<span>57x^5/19x^2
= 3x^3

hope it helps</span>
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4. A salesman has a basic salary and, in addition, recelves a commission which is a fixed percentage of his sales volume, when h
vekshin1

The commission is 20% of sales, the basic salary is $200 and the equation is y = 0.2x + 200

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Let y represent the total sales for x weekly sales, m represent the commision and b represent the basic salary. Hence:

400 = 1000m + b  (1)

Also:

300 = 500m + b  (2)

Hence:

m = 0.2 = 20%, b = $200

The commission is 20% of sales, the basic salary is $200 and the equation is y = 0.2x + 200

Find out more on equation at: brainly.com/question/2972832

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2 years ago
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Answer:

HAPPY BURFDAY

Step-by-step explanation:

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You want to put a 5 inch thick layer of topsoil for a new 23 ft by 18 ft garden. the dirt store sells by the cubic yards. how ma
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5 inches = 5/12 = 0.4166 feet

23 x 18 x 0.4166 = 172.5 cubic feet

1 cubic foot = 0.037037 cubic yards

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John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
3 years ago
7. The probability that a randomly chosen student will be left-handed is .09: a) In a class of 108, find the probability that th
vampirchik [111]

Solution :

Let x be student will be left handed

P = 0.09

Using the normal approximation to binomial distribution,

a). n = 108,

    μ = np = 9.72

    $\sigma = \sqrt{np(1-p)}$

       $=\sqrt{8.8452}$

       = 2.9741

Required probability,

P(x=8) = P(7.5 < x < 8.5)

$=P\left(\frac{7.5-9.72}{2.9741}< \frac{x-\mu}{\sigma}< \frac{8.5-9.72}{2.9741}\right)$

$=P(-0.75 < z < -0.41)$

Using z table,

= P(z<-0.41)-P(z<-0.75)

= 0.3409-0.2266

= 0.1143

b). P(x=12) = P(11.5 < x < 12.5)

$=P\left(\frac{11.5-9.72}{2.9741}< \frac{x-\mu}{\sigma}< \frac{12.5-9.72}{2.9741}\right)$

$=P(0.60 < z < 0.94)$

Using z table,

= P(z< 0.94)-P(z< 0.60)

= 0.8294 - 0.7257

= 0.1006

7 0
3 years ago
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