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dimaraw [331]
3 years ago
13

Drag the tiles to classify each number as rational or irrational. Each tile may be used more than once.

Mathematics
1 answer:
shtirl [24]3 years ago
6 0

Answer:

Step-by-step explanation:

1. rational

2.inrational

3.in

4. rational

5. in

edge 2021

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4x-y=2 2x+9y=-12 unsure how to solve
Anna11 [10]
Simple, use the process of elimination to figure this out:

4x-y=2
2x+9y=-12

Convert these, to be able to find either x or y.

2(4x-y=2)   8x-2y=4   multiply by 2.
then
4(2x+9y=-12)  8x+36y=-48

Which makes them

8x-2y=4
-
8x+36y=-48

-38y=52

This simplified makes it y= -26/19.

Then plug y back into the original equation

4x-y=2

4x-(-26/19)=2

Making it x= 3/19.

Thus your Point of Intersection is at (3/19, -26/19)
8 0
4 years ago
Read 2 more answers
Can I get some help please?
Travka [436]

Answer and Step-by-step explanation:

A) The first three terms are:

5, 8, 11

B) The tenth term is 32

You find the answer by plugging in the value for n. So for A), you would plug in 1, 2, and 3 (the first 3 terms), then to find the tenth term, you plug in 10 for n.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

3 0
3 years ago
A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a
Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

                                   p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

p^4 (1-p)

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

(1-p)^4p

Therefore the probability that you stop at the fifth flip would be

                                    p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

7 0
3 years ago
Hi! I need help on this, make sure to label each part with A,B,C, and D so I don’t type it in wrong. I will fine brainlist
ivolga24 [154]

Answer:

Step-by-step explanation:

uh..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

7 0
3 years ago
Dave collects rocks. he makes 12 groups of 10 rocks and has none left over. How many rocks does Dave have?
Firdavs [7]
That means he has 120 rocks
12*10=120

4 0
4 years ago
Read 2 more answers
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