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3 years ago

What is 7 and 8? A and B

Mathematics

2 answers:

victus00 [196]3 years ago

6 0

Yes I agree!!! :D hope u get it correct in class or what ever

Lerok [7]3 years ago

3 0

7. For a triangle to be equilateral, all 3 sides must be of equal length. This implies that all corners must be equal. To find out if this is the case we have to find variable x. Never draw conclusions from the drawing.

We know that all corners of the triangle should add up to 180. So:

2x + 2x + (3x-30) = 180 => 7x = 210 => x=30

Also, 2x = 3x-30. All cornders are 60°!

So yes, this triangle is indeed equilateral. If it is equilateral, it is for sure also isosceles (for that, only two sides need to be the same).

8.A circumcenter
8.B incenter

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What is a tenth of a decimal ?

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One decimal place to the left decimal point is the place.

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3 years ago

Simplify (3^3)(1/9^2 − 6 + 2)

AnnZ [28]

Answer:

27/77

Step-by-step explanation:

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2 years ago

Composition about myself in 150 words

tigry1 [53]

Answer:

I am a student who’s studying in a prestigious college in Bangalore city. It’s a city where I grew up. I live in a town with my family. The school I studied till 12th is also in the town.

Things I am good at

Almost everyone is kind at least one sport. The one competition that I am good at is basketball. In my school, almost everyone had an obsession with the sport, and so did I. Every game period, my teacher would make us play basketball, along with other games. Over the years, the way I played basketball improved, and while learning the game, I discovered other lessons as well. One of the lessons I’ve learned is how to play in a team. When you play in a group, you depend on each other for winning.

I have always been energetic and lively. While many people feel awkward and weird, making me friends, I have no problems with making new friends. I can talk to everyone quickly and know them.

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8 0

2 years ago

The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

2 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

3 0

2 years ago