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2 years ago

How to solve the equation for x/12+3

2 answers:

7 0

X= -36 because I subtracted 3 from 0 and the equation should now look like this: x/12 = -3. From there you solve by multiplying the reciprocal of x/12 and that would be 12/1. Doing this just cancels the 12 and multiplies the -3 by 12 so now you should be left with X= -36

Thepotemich [5.8K]2 years ago

3 0

Answer:

gggGG

Step-by-step explanation:

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Which of the following does not describe slope?

kolbaska11 [484]

<span>a. A two-dimensional figure</span>

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3 years ago

Write annequivalent fraction by multiplying 2/3

STatiana [176]

$\bf \cfrac{2}{3}\cdot \cfrac{same}{same}\implies \stackrel{equivalent~fraction}{\cfrac{2(same)}{3(same)}}\qquad therefore\qquad \cfrac{2}{3}\cdot \cfrac{9}{9}\implies \stackrel{equivalent}{\cfrac{18}{27}}$

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3 years ago

A polymer is manufactured in a batch chemical process. Viscosity measurements show that it is approximately normally distributed

PilotLPTM [1.2K]

Answer: (751.05, 766.95)

Step-by-step explanation:

We know that the confidence interval for population mean is given by :-

$\overline{x}\pm z*\dfrac{\sigma}{\sqrt{n}}$ ,

where $\sigma$ =population standard deviation.

$\overline{x}$ = sample mean

n= sample size

z* = Two-tailed critical z-value.

Given : $\sigma= 20$

n= 42

$\overline{x}=759$

We know that from z-table , the two-tailed critical value for 99% confidence interval : z* =2.576

Now, the 99% confidence interval around the true population mean viscosity :-

$759\pm (2.5760)\dfrac{20}{\sqrt{42}}\\\\=759\pm (2.5760)(3.086067)\\\\=759\pm7.9497=(759-7.9497,\ 759+7.9497)\]\\=(751.0503,\ 766.9497)\approx(751.05,\ 766.95)$

∴ A 99% confidence interval around the true population mean viscosity : (751.05, 766.95)

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3 years ago

PLEASE ANSWER THIS QUESTION !! 20 POINTS AND BRAINLIEST !!

lutik1710 [3]

Answer:

Step-by-step explanation:

If they have no solutions, the lines must not intersect.

If they don't intersect, they must be parallel

8 0

2 years ago

Read 2 more answers

list the pair of congruent angles and the extended proportion that relations the corresponding sides for the similar polygons KL

kherson [118]

Answer:

In Polygons KLMNP and HGFDC,

Angles K, L, M, N and P are corresponding to H, G, F, D and C,

And, Sides KL, LM, MN, NP and PK are corresponding to sides HG, GF, FD, DC and CH

If the two figures are similar then their corresponding sides and angles are congruent.

Therefore, If Polygon KLMNP is similar to HGFDC,

Then, ∠ K ≅ ∠ H, ∠ L ≅ ∠ G, ∠ M ≅ ∠ F, ∠ N ≅ ∠ D, ∠ P ≅ ∠ C

And, KL ≅ HG, LM ≅ GF, MN≅FD, NP≅DC, PK≅CH

7 0

2 years ago