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Artyom0805 [142]
3 years ago
6

Shape of distribution

Mathematics
1 answer:
qwelly [4]3 years ago
5 0

Answer:

where is the picture so i can help you

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Explain how using basic facts can help you find 10x20x30x40 mentally.
sladkih [1.3K]
You know that anything times a one is that number, so multiply 2,3, and 4, and then multiply that by one. once you get the answer, put the zeros behind each multiplied number behind your answer, and there you go.

2×3×4= 24×1=24
there are 4 zeros
240,000
4 0
3 years ago
You play the following game against your friend. You have 2 urns and 4 balls One of the balls is black and the other 3 are white
Rom4ik [11]

Answer:

Part a: <em>The case in such a way that the chances are minimized so the case is where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b: <em>The case in such a way that the chances are maximized so the case  where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c: <em>The minimum and maximum probabilities of winning  for n number of balls are  such that </em>

  • <em>when all the n balls are placed in one of the urns the probability of the winning will be least as 1/2n</em>
  • <em>when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, as 0.5</em>

Step-by-step explanation:

Let us suppose there are two urns A and A'. The event of selecting a urn is given as A thus the probability of this is given as

P(A)=P(A')=0.5

Now the probability of finding the black ball is given as

P(B)=P(B∩A)+P(P(B∩A')

P(B)=(P(B|A)P(A))+(P(B|A')P(A'))

Now there can be four cases as follows

Case 1: When all the four balls are in urn A and no ball is in urn A'

so

P(B|A)=0.25 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.25*0.5)+(0*0.5)

P(B)=0.125;

Case 2: When the black ball is in urn A and 3 white balls are in urn A'

so

P(B|A)=1.0 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1*0.5)+(0*0.5)

P(B)=0.5;

Case 3: When there is 1 black ball  and 1 white ball in urn A and 2 white balls are in urn A'

so

P(B|A)=0.5 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.5*0.5)+(0*0.5)

P(B)=0.25;

Case 4: When there is 1 black ball  and 2 white balls in urn A and 1 white ball are in urn A'

so

P(B|A)=0.33 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.33*0.5)+(0*0.5)

P(B)=0.165;

Part a:

<em>As it says the case in such a way that the chances are minimized so the case is case 1 where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b:

<em>As it says the case in such a way that the chances are maximized so the case is case 2 where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c:

The minimum and maximum probabilities of winning  for n number of balls are  such that

  • when all the n balls are placed in one of the urns the probability of the winning will be least given as

P(B|A)=1/n and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/n*1/2)+(0*0.5)

P(B)=1/2n;

  • when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, equal to calculated above and is given as

P(B|A)=1/1 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/1*1/2)+(0*0.5)

P(B)=0.5;

5 0
3 years ago
A. In the Example, what is<br> 3/4 / 1/3<br> the unit rate for?
hoa [83]

Answer:

the unit rate for it would be 1/4

6 0
1 year ago
1. Exercise 18, section 5.4. Compute 90, 91, 92, 93, 94 and 95. Make a conjecture about the units digit of 9n where n is a posit
Burka [1]

Answer:

n is a positive integer and here 90, 91, 92, 93, 94 and 95 is 9⁰ = 1

9¹ = 9

9² = 81

9³ = 729

9⁴ = 6561

9⁵ = 59049

so mathematical induction is prove given below.

Step-by-step explanation:

9⁰ = 1

9¹ = 9

9² = 81

9³ = 729

9⁴ = 6561

9⁵ = 59049

lets us consider by P (n) for the base case n = 0 then 9⁰ = 1  So P (1) is true

k ≥ 1so consider all integers

1≤l≤ k

There is a need to prove P (k+1)

if k is odd

9^k^+^1 =9^k.9.

9^k is 9 so

9^k = 9 +10m

then

9^k^+^1 = 9 ( 9 + 10m)

       =81 + 10 (9m)

so digit is 1 and k +1 is even

3 0
3 years ago
X=?? solve for x hurry!
jarptica [38.1K]

Answer: x + 30 + 90= 180(sum of angles on a straight line)

x= 180-120

x=60

7 0
3 years ago
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