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3 years ago

What is Show your work

Mathematics

1 answer:

zloy xaker [14]3 years ago

5 0

Multiplying both sides of this equation by 3, to eliminate the fraction, we get:

9 + m = 6.

Next, we subtract 9 from both sides to isolate m: m = 6 - 9 = -3

Then m = -3.

Check: Subst. -3 for m in this equation:

9 - 3 6

------- = ------ = 2 (as expected). Thus, m = -3 is correct.

3 3

Problem #10:

4n - 9 = -9 Add 9 to both sides: 4n = -9 + 9 = 0.

Divide both sides by 4: n = 0.

Check: Substitute 0 for n: 4(0) - 9 = -9 is true.

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The discriminant of quadrilateral equation X square + X + 0<br>is dash

grandymaker [24]

Answer:

Step-by-step explanation:

D = b²- 4 ac

1²- 4(1)(0)

= 1-0

7 0

3 years ago

A certain firm has plants a, b, and c producing respectively 35\%, 15\%, and 50\% of the total output. The probabilities of a no

TEA [102]

The proportion of production that is defective and from plant A is

... 0.35·0.25 = 0.0875

The proportion of production that is defective and from plant B is

... 0.15·0.05 = 0.0075

The proportion of production that is defective and from plant C is

... 0.50·0.15 = 0.075

Thus, the proportion of defective product that is from plant C is

... 0.075/(0.0875 +0.0075 +0.075) = 75/170 = 15/34 ≈ 44.12%

_____

P(C | defective) = P(C&defective)/P(defective)

8 0

3 years ago

588<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B%3F%5D%7B%3F%7D%20" id="TexFormula1" title=" \sqrt[?]{?} " alt=" \sqrt[?]{

zaharov [31]

Answer:

660 :)

Step-by-step explanation:

Just simple math

4 0

3 years ago

The mean computed from ungrouped data is a more accurate measure than the mean computed from grouped data?

Rufina [12.5K]

It doesn’t make any difference whether or not the data is grouped, the mean is the same. So b. false.

4 0

3 years ago

X2+y2-10x-16y+53=0 what is the center and radius of the circle ?

masha68 [24]

The center-radius form of the circle equation

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have:

$x^2+y^2-10x-16y+53=0$

Use $(a-b)^2=a^2-2ab+b^2\qquad(*)$

$x^2-10x+y^2-16y+53=0\qquad\text{subtract 53 from both sides}\\\\x^2-2(x)(5)+y^2-2(y)(8)=-53\qquad\text{add}\ 5^2\ \text{and}\ 8^2\ \text{to both sides}\\\\\underbrace{x^2-2(x)(5)+5^2}_{(*)}+\underbrace{y^2-2(y)(8)+8^2}_{(*)}=5^2+8^2-53\\\\(x-5)^2+(y-8)^2=25+64-53\\\\(x-5)^2+(y-8)^2=36\\\\(x-5)^2+(y-8)^2=6^2\\\\Answer:\\\\\boxed{center:(5,\ 8)}\\\\\boxed{radius:r=6}$

4 0

3 years ago