<img src="https://tex.z-dn.net/?f=6%2Bx%3D21" id="TexFormula1" title="6+x=21" alt="6+x=21" align="absmiddle" class="latex-formul

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Digiron [165]

2 years ago

a">
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Mathematics

2 answers:

Brrunno [24]2 years ago

8 0

<h2>Answer:</h2>

<h2>Step-by-step explanation:</h2>

$6 + x = 21 \\ \\ x = 21 - 6 \\ \\ x = 15$

babymother [125]2 years ago

4 0

21 - 6 = x

so the answers is 15

21 - 6 = 15

check:
15 + 6 = 21

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A system of inequalities is graphed below. The variable x represents the number of cans someone is buying, and y represents the

8090 [49]

We have that
<span>We will analyze each case to verify the answer
see the attached figure

</span>Point A) 7 cans and 140 bottles------------> (x,y)=(7,140)----> is not solution
Point B) 12 cans and 40 bottles------------> (x,y)=(12,40)----> is not solution
Point C) 8 cans and 120 bottles------------> (x,y)=(8,120)----> is a solution
Point D) 9 cans and 70 bottles------------> (x,y)=(9,70)-------> is a solution

the answer is
the solutions are
<span>8 cans and 120 bottles
9 cans and 70 bottles</span>

6 0

3 years ago

سا (a) From the definition of derivatives determine dy÷dx if y = -2÷x

harina [27]

Step-by-step explanation:

Given: $y = -\dfrac{2}{x}$

Derivative of a power function $x^n$ :

$\dfrac{d}{dx}(x^n) = nx^{n-1}$

Therefore,

$\dfrac{dy}{dx}=-2(-1)x^{-2} = \dfrac{2}{x^2}$

3 0

3 years ago

15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.

Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = $30C5 = \frac{30 !}{(30-5)!5!} \\$

$30C5 = 142506$ ways

Number of ways of selecting 5 good bulbs from 26 bulbs = $26C5 = \frac{26 !}{(26-5)!5!} \\$

$26C5 = 65780$ ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = $\frac{26C4 * 4C1}{30C5}$

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) = $\frac{26C3 * 4C2}{30C5}$

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) = $\frac{26C2 * 4C3}{30C5}$

Pr(3 defective) = 0.009

Pr(4 defective) = $\frac{26C1 * 4C4}{30C5}$

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0

3 years ago

Is there a proportional relationship between x and y? Explain.

Vlad1618 [11]

I don’t think so but divide the top one by the bottom one for every one of them and if it is all the same answer they are proportional x/y

4 0

2 years ago

Bro please help for a brainlist for best answer !!!!!!

tatiyna

The domain of a function are the possible input values of the function.

<em>The domain of the function is (d) Whole numbers</em>

First, we identify the start and end values of x on the graph.

We have:

$Start = 0$

The graph has no end; so:

$End = \infty$

This means that the domain of the graph is $(0,\infty)$

The number of times one can visit the pool cannot be decimal.

Hence, the domain of the function is whole numbers