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4 years ago

Determine the relation of the following lines y = -2x +4 and y= -2x +1

Mathematics

1 answer:

Ivahew [28]4 years ago

7 0

They're parallel because they have the same slope.

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What is 1,680+110 7/16

natita [175]

I don't get because you did not add a symbol between 110 and 7/16 so i can't tell you.

5 0

4 years ago

Tom mows lawns, after mowing 5 lawns he made 100$. Assuming he

BARSIC [14]

Answer:

$20

Step-by-step explanation:

simple maths

5 0

3 years ago

How to write a paragraph proof? Given: ∠T and ∠V are right angles TW║VU.<br> Prove: ΔTUW≅ΔVWU

mafiozo [28]

Note: The missing diagram is attached below for a reference.

Answer:

Δ $TUW$ ≅ Δ $VWU$ by $AAS$ case.

Step-by-step explanation:

Given:

∠T and ∠V are right angles.

$TW$ ║ $VU$

To Prove:

Δ $TUW$ ≅ Δ $VWU$

∠ $T$ ≅ ∠ $V$ ≅ $90^{0}$ (Given)

Side $UW$ ≅

$TW$ ║ $UV$

and $UW$ is a transverse

So,

∠ $TWU$ ≅ ∠ $WUV$ $\left[alternate\:\:interior\:\:angles\right]$

$As\:Angle\:=\:Angle\:=\:side\:are\:equal$

Thus,

Δ $TUW$ ≅ Δ $VWU$

Keywords: proof, paragraph, logic, reason

Learn more about proof and reason from brainly.com/question/13396866

#learnwithBrainly

4 0

3 years ago

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo

muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: $P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}$

Where $\lambda$ is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

$P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}$

Substitute the respective values in the above formula.

$P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})$

$P(k > 2)= 1 - (0.44596)$

$P(k > 2)=0.55404$

Hence, the required probability is 0.55404.

4 0

3 years ago

Help with a b and c

djyliett [7]

5/6 x 2/2 = 10/12
1/4 x 3/3 = 3/12
2/3 x 4/4 = 8/12

5/6 + 1/4 = 13/12 (over 1 yard)
1/4 + 2/3 = 11/12 (slightly less than 1 yard)
5/6 + 2/3 = 18/12 (too much)

Here is the answer for a:
a) 5/6 and 1/4

5 0

4 years ago