Answer:
total no of outcomes are '15'
outcomes are shown below.
Step-by-step explanation:
A = {1, 2, 3, 4, 5 ,6 ,7}
subsets which will include {1, 3, 5}
hence, the outcome are :
{1,2,3,5}, {1,3,4,5}, {1,3,5,6}, {1,3,5,7}, {1,2,3,5,6},{1,2,3,5,7},{1,2,3,4,5}, {1,3,4,5,6}, {1,3,4,5,7}, {1,2,3,5,7},{1,2,3,5,6}, {1,3,5,6,7},{1,2,3,4,5,6},{1,2,3,4,5,7}, {1,2,3,4,5,6,7}
so the total no of outcomes are '15'
Answer:
The sample size needed if the margin of error of the confidence interval is to be no more than 0.02.
n= 2015
Step-by-step explanation:
Given the customer relations department will survey a random sample of customers and compute a 95% confidence interval for the proportion that are dissatisfied
Given data from past studies, they believe that this proportion will be about 0.28
The proportion of success(p) = 0.28
We know that the margin of error of 95% of intervals of given proportion
margin of error = …(i)
Given margin of error = 0.02
Substitute values in equation (i) cross multiplication √n
0.02 √n = 2√0.28X0.72
On calculation, we get √n = 44.89
squaring on both sides, we get
n = 2015
Conclusion:-
The sample size needed if the margin of error of the confidence interval is to be no more than 0.02.
n= 2015
What do I need to d
Answer: 219,185
-Multiply the place values to find the answer,
Answer:
7+1x=8, 8+10 =x18 is your answer
1/-4^-5
Answer: I got -1024...
Step-by-step explanation:
Answer: The expected number of spades that you will draw is 0.751 spades
Step-by-step explanation:
The expected value can be calculated as:
∑xₙ*pₙ
Where xₙ is the n-th event, and pₙ is the probability of that event.
First, let's count the possible events and calculate the probability for each one.
x₀ = drawing 0 spades.
Out of 52 cards, we have only 13 spades, then 52 - 13 = 39 are not spades.
Then the probability of not drawing a spade in the first draw is:
p1 = 39/52
In the second draw we will have a card less than before in the deck (so we have 38 cards that are not spades, and 51 cards in total), then the probability of not drawing a spade is:
p2 = 38/51
And with the same reasoning, in the third draw the probability is:
p3 = 37/50
The joint probability for this event will be:
p₀ = p1*p2*p3 = (39/52)*(38/51)*(37/50) = 0.413
Second event:
x₁ = drawing one spade.
Let's suppose that in the first draw we get the spade, the probability will be:
p1 = 13/52
In the second draw, we get no spade, then the probability is:
p2 = 39/51
in the third draw we also get no spade, the probability is:
p3 = 38/50
And we also have the case where the spade is drawn in the second draw, and in the third draw, then we have 3 permutations, this means that the probability of drawing only one spade is:
p₁ = 3*p1*p2*p3 = 3*(13/52)(39/51)*(38/50) = 0.436
third event:
x₂ = drawing two spades:
Let's assume that in the first draw we do not get a spade, then the probabilities are:
p1 = 39/52
p2 = 13/51
p3 = 12/50
And same as before, we will have 3 permutations, because we could not draw a spade in the second draw, or in the third, then the probability for this case is:
p₂ = 3*p1*p2*p3 = 3*( 39/52)*(13/51)*(12/50) = 0.138
And the last event:
x₃ = drawing 3 spades.
The probabilities will be:
p1 = 13/52
p2 = 12/51
p3 = 11/50
And there are no permutations here, so the joint probability is:
p₃ = p1*p2*p3 = (13/52)*(12/51)*(11/50) = 0.013
Now we can calculate the expected value:
EV = 0*0.413 + 1*0.436 + 2*0.138 + 3*0.013 = 0.751
The expected number of spades that you will draw is 0.751 spades
The expected number of spades drawn when drawing three cards without replacement from a standard deck is approximately 0.75 spades.
To calculate this, we can use the concept of conditional probability. Initially, there are 13 spades out of 52 cards in the deck, giving us a 13/52 chance of drawing a spade on the first card.
If the first card drawn is a spade, there are now 12 spades left out of 51 cards, so the probability of drawing a spade on the second card is 12/51.
If the first two cards are spades, there are 11 spades left out of 50 cards for the third draw, with a probability of 11/50.
Now, we multiply these probabilities together and sum up the possible scenarios (0, 1, 2, or 3 spades drawn) to get the expected value: (0 * (39/52 * 38/51 * 37/50)) + (1 * (13/52 * 39/51 * 38/50 + 39/52 * 12/51 * 38/50 + 39/52 * 38/51 * 11/50)) + (2 * (13/52 * 12/51 * 39/50 + 13/52 * 39/51 * 11/50 + 39/52 * 12/51 * 11/50)) + (3 * (13/52 * 12/51 * 11/50)) ≈ 0.75 spades.
So, the expected number of spades drawn when selecting three cards without replacement from a standard deck is approximately 0.75.
This means, on average, you can expect to draw about 3/4 of a spade.
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Answer:
0.288
Step-by-step explanation:
Given that :
Correlation (R) = 0.48
Slope of linear model which predicts Lifespan from years of education (m) = 0.8
To determine the value of slope of the model which predicts years of eductauoon from lifespan:
The square of the regression Coefficient is multiplied by the inverse of the slope of linear model which predicts Lifespan from years of education
Hence,
(R² * 1/m)
0.48² * 1/0.8
0.2304 * 1.25
= 0.288
The slope of the line that predicts years of education from lifespan is 1.25.
The slope of the line that predicts years of education from lifespan can be determined by taking the reciprocal of the slope that predicts lifespan from years of education. In this case, the slope of the line that predicts years of education from lifespan would be 1 divided by 0.8, which equals 1.25.
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