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erik [133]
3 years ago
11

PLEASE HELP ME ANSWER THIS IT WILL HELP A LOT THANK YOU ​

Mathematics
2 answers:
PilotLPTM [1.2K]3 years ago
8 0
<h3>Kindly refer to the attachment above!</h3>

USPshnik [31]3 years ago
6 0

\\ \sf\longmapsto \dfrac{6x^2y^3}{(-4x^4y)(-3x^3y^2)}

\\ \sf\longmapsto \dfrac{6x^2y^3}{12x^{4+3}y^{2+1}}

\\ \sf\longmapsto \dfrac{6x^2y^3}{12x^7y^3}

\\ \sf\longmapsto \dfrac{1}{2}x^{2-7}y^{3-3}

\\ \sf\longmapsto \dfrac{1}{2}x^{-5}y^0

\\ \sf\longmapsto \dfrac{1}{2x^5}

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Fawn plants 2/3 of the garden with vegetables. Her son plants the remainder of the garden. He decides to use 1/2 of his plant fl
bulgar [2K]

Given that Fawn plants vegetables into garden = \frac{2}{3}

Exact amount of garden is not given so we can assume that amount of garden is 1.

Then remaining amount of garden =1-\frac{2}{3}=\frac{3}{3}-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}

Now half of the remaining part of garden that is half of \frac{1}{3} rd part of garden is planted with flowers

so that will be \frac{1}{2}*\frac{1}{3}=\frac{1}{6}

Hence required fraction for the answer will be 1/6.

3 0
3 years ago
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What is the domain of x = 4?​
34kurt
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined. The range is the set of all valid y values.
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3 years ago
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Newton’s first law of motion, please show work if you can :)
aniked [119]

Answer:

8kg

Step-by-step explanation:

Force = Mass × Acceleration.

Where:

Force= 20N

Acceleration= 2.5m/s^2

Mass = ?

20 = m × 2.5

20= 2.5m

20/2.5 = m

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5 0
3 years ago
(a) Consider a class with 30 students. Compute the probability that at least two of them have their birthdays on the same day. (
Galina-37 [17]

Answer:

a.) 0.7063

b.) 23

Step-by-step explanation:

a.)

Let X be an event in which at least 2 students have same birthday

     Y be an event in which no student have same birthday.

Now,

P(X) + P(Y) = 1

⇒P(X) = 1 - P(Y)

as we know that,

Probability of no one has birthday on same day = P(Y)

⇒P(Y) = \frac{365!}{(365)^{n} (365-n)! }      where there are n people in a group

As given,

n = 30

⇒P(Y) = \frac{365!}{(365)^{30} (365-30)! } = \frac{365!}{(365)^{30} (335)! } = 0.2937

∴ we get

P(X) = 1 - 0.2937 = 0.7063

So,

The probability that at least two of them have their birthdays on the same day  =  0.7063

b.)

Given, P(X) > 0.5

As

P(X) + P(Y) = 1

⇒P(Y) ≤ 0.5

As

P(Y) = \frac{365!}{(365)^{n} (365-n)! }

We use hit and trial method

If n = 1 , then

P(Y) = \frac{365!}{(365)^{1} (365-1)! } = \frac{365!}{(365)^{1} (364)! }  = 1 \nleq 0.5

If n = 5 , then

P(Y) = \frac{365!}{(365)^{5} (365-5)! } = \frac{365!}{(365)^{5} (360)! }  = 0.97 \nleq 0.5

If n = 10 , then

P(Y) = \frac{365!}{(365)^{10} (365-10)! } = \frac{365!}{(365)^{10} (354)! }  = 0.88 \nleq 0.5

If n = 15 , then

P(Y) = \frac{365!}{(365)^{15} (365-15)! } = \frac{365!}{(365)^{15} (350)! }  = 0.75 \nleq 0.5

If n = 20 , then

P(Y) = \frac{365!}{(365)^{20} (365-20)! } = \frac{365!}{(365)^{20} (345)! }  = 0.588 \nleq 0.5

If n = 22 , then

P(Y) = \frac{365!}{(365)^{22} (365-22)! } = \frac{365!}{(365)^{22} (343)! }  = 0.52 \nleq 0.5

If n = 23 , then

P(Y) = \frac{365!}{(365)^{23} (365-23)! } = \frac{365!}{(365)^{23} (342)! }  = 0.49 \nleq 0.5

∴ we get

Number of students should be in class in order to have this probability above 0.5 = 23

5 0
3 years ago
The triangles shown below must be congruent.
VashaNatasha [74]

Answer:

it is true

Step-by-step explanation:

If two angles and the non-included side of one triangle are equal to the corresponding angles and side of another triangle, the triangles are congruent.

7 0
4 years ago
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