Answer:
Actually, what you said you have so far is not correct. The 2 correct answers are the 1st one (x + y = 15) and the 5th one (15x + 10y > 180)
Step-by-step explanation:
If tutoring French is x hours and scooping ice cream is y hours and he is going to work 15 hours for sure doing both, then we can add them together to get that x hours + y hours = 15 hours, or put simply: x + y = 15.
Now we are going to throw in the added fun of the money he makes doing each. The thing to realize here is that we can only add like terms. So looking at the equation above, we have x hours of tutoring and y hours of scooping, so if we want to add them, we will add those number of hours together to get the total number of hours he worked, which we know to be 15. The same goes for money. If we add money earned from tutoring to money earned from scooping, we need that to be greater than the money he wants to earn which is 180 at least. Because he wants to earn MORE than $180. we use the ">" sign. Since he earns $15 an hour tutoring, that expression is $15x; since he earns $10 an hour scooping, that expression is $10y. Now add them together (and you CAN because they are both expressions relating dollars to dollars) and set the sum > $180:
$15x + $10y > $180. That's why your answer is not correct. Use mine (with the understanding that you care about why yours is wrong and mine is correct) and you'll be fine.
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:
Your answer is C
Step-by-step explanation:
Supplementary angles are angles that come out to 180 degrees when added.
90+90=180
The curve representing the relationship includes the point
The growth factor is
An earthquake with an energy level of has a magnitude of
The equation 'M = 0.6666 log x - 3.2' describes the relationship between the energy produced by an earthquake (x) and its resulting magnitude (M). The 0.6666 in the formula is the growth factor. To get a magnitude for a specific energy level, we plug the energy value into the formula, and solve.
The equation given represents the relationship between the magnitude of an earthquake (M) and the energy it produces (x), with M being the earthquake's magnitude and x being the energy produced by the earthquake. This equation is a logarithmic model with a base of 10 (without the base explicitly stated, we assume it to be 10).
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Answer:
Step-by-step explanation:
The probability that it will take more than 6 minutes for all the customers in line to check out is 0.40.
We are given the probability distribution of x, the number of customers in line at a supermarket express checkout counter.
Moreover, we are given that each customer takes 3 minutes to check out.
It means that if there are 0 customers in line, i.e., x=0, then it will take 0 minutes for all the customers currently in line to check out.
If there is 1 customer in line, i.e., x=1, then it will take 3 minutes for all the customers currently in line to check out.
If there are 2 customers in line, i.e., x=2, then it will take 6 minutes for all the customers currently in line to check out.
If there are 3 customers in line, i.e., x=3, then it will take 9 minutes for all the customers currently in line to check out.
If there are 4 customers in line, i.e., x=4, then it will take 12 minutes for all the customers currently in line to check out.
If there are 5 customers in line, i.e., x=5, then it will take 15 minutes for all the customers currently in line to check out.
From above we note that if there are 3 or more customers in the line, then it will take more than 6 minutes (note that the case of check out time equal to 6 minutes is not included when we want 'more than 6 minutes') for all the customers currently in line to check out.
Thus, required probability is given by:
P(more than 6 minutes for all the customers currently in line to check out) = P(x ≥ 3)
= P(x=3) + P(x=4) + P(x=5)
= 0.20 + 0.15 + 0.05
= 0.40
Therefore, the probability that it will take more than 6 minutes for all the customers in line to check out is 0.40.
To learn more about the probability visit:
brainly.com/question/11234923.
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Without specific information on the total number of customers or the distribution of customers in line, we cannot calculate a specific probability for it to take more than 6 minutes for all customers to check out, given that each customer takes 3 minutes.
The question is about the probability that it will take more than 6 minutes for all the customers in line to check out, given that each customer takes 3 minutes. The time it takes for all the customers to check out is determined by the number of customers in line. If there are two or more customers in line, it will definitely take more than 6 minutes for all of them to check out, because the checkout time is 3 minutes per customer.
So, the question of probability relates to the likelihood of there being two or more customers in line. Without information on the total number of customers, or the distribution of customers in line, we cannot calculate a specific probability.
Please note, this is a practical application of topics in probability and queue theory, involving concepts like mean arrival rate and service rate.
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The measure angle of 4?