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Dennis_Churaev [7]
2 years ago
6

Anthony borrows $6.50 from a friend to pay for a sandwich and drink at lunch. Then he borrows $3.75 from his sister to pay a lib

rary fine. Choose an equation to express Anthony's total debt that day.
Mathematics
3 answers:
Katen [24]2 years ago
8 0

Money borrowed by Anthony from his friend for a sandwich and drink at lunch = $6.50

Money borrowed by Anthony from his sister to pay a library fine = $3.75

Hence, the equation to express Anthony's total debt that day will be =

Let T be the total debt , so the equations becomes,

6.50 + 3.75 = T

damaskus [11]2 years ago
8 0

Answer:

-\$6.50+(-\$3.75)=-\$10.25

Step-by-step explanation:

We have been given that Anthony borrows $6.50 from a friend to pay for a sandwich and drink at lunch. Then he borrows $3.75 from his sister to pay a library fine.

Anthony borrowed money, so a negative will represent his debt.

\text{Anthony's total debt}=-\$6.50+(-\$3.75)

-\$10.25=-\$6.50+(-\$3.75)

Therefore, our required equation would be -\$6.50+(-\$3.75)=-\$10.25

Guest1 year ago
0 0

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Answer:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

Step-by-step explanation:

For the random variable N_1 we define the possible values for this variable on this case [1,2,3,4,5] . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable N_1 like this:

P(N_1 = a) = \frac{5-a C 1}{5C2}

For the distribution of N_2 we need to take in count that we are finding a conditional distribution. N_2 given N_1 =a, for this case we see that N_2 \in [1,2,...,5-a], so then exist 5-a C 1 ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}

And if we want to find the joint probability we just need to do this:

P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)

And if we multiply the probabilities founded we got:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

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- The ball is thrown upward with initial velocity 48 feet/second

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Another way to do this:

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