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sineoko [7]
3 years ago
8

Six friends each het 3/8 of a pound of candy. how much candy will the friends get in all

Mathematics
1 answer:
tamaranim1 [39]3 years ago
6 0
6(3/8)

18/8

9/4

2 1/4 or

2.25 lbs
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defon
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
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Do you have the rest of the question?
6 0
3 years ago
Jimmy ran at a speed of m miles. how far did he run in t minutes
beks73 [17]

Answer:

(mxt) miles

Step-by-step explanation:

speed = m \: miles \: per \:minute \\ time \:  = t \: minutes \\ speed =  \frac{distance}{time}  \\ m =  \frac{d}{t}  \\ m \times t \:  = d \\ d = (m \times t)miles

hope that this is helpful.

8 0
2 years ago
PLEASE HELP!! DESPERATE
9966 [12]

a) i) Since blood types across individuals are independent, we have

Pr[1 = A and 2 = B] = Pr[1 = A] • Pr[2 = B]

where e.g. Pr[1 = A] means "probability that partner 1 has type A blood". According to the given distribution,

Pr[1 = A and 2 = B] = 0.37 • 0.13 = 0.0481 = 4.81%

a) ii) The event that partner 1 has type A and partner 2 has type B is exclusive from the event that partner 1 has type B and partner 2 has type A, and vice versa. In other words, only one of these events can happen, and not both simultaneously. This means the probability of either event occurring is equal to the sum of the probabilities of these events occurring individually. Then

Pr[(1 = A and 2 = B) or (1 = B and 2 = A)]

= Pr[1 = A and 2 = B] + Pr[1 = B and 2 = A]

Both probabilities are the same, and we found it in the previous part. So

Pr[(1 = A and 2 = B) or (1 = B and 2 = A)]

= 0.0481 + 0.0481 = 0.0962 = 9.62%

a) iii) The event that at least one partner has type O occurs if exactly one partner has type O, or both do. That is, "at least one partner with type O" contains 3 possible events,

• 1 = O and 2 = not O

• 1 = not O and 2 = O

• 1 = O and 2 = O

For the first event, by independence, is

Pr[1 = O and 2 = not O] = Pr[1 = O] • Pr[2 = not O]

and by definition of complementary event,

Pr[1 = O and 2 = not O] = Pr[1 = O] • (1 - Pr[2 = O])

From the given distribution, it follows that

Pr[1 = O and 2 = not O] = 0.44 • (1 - 0.44) = 0.2464 = 24.64%

The second event is the same as the first in terms of probability.

For the third event, we have

Pr[1 = O and 2 = O] = Pr[1 = O] • Pr[2 = O] = 0.44² = 0.1936 = 19.36%

Then the total probability of at least one O-type partner is

24.64% + 24.64% + 19.36% = 68.64%

b) Suppose partner 1 is type-B. Then partner 2 is an acceptable donor if they are either type-B or type-O. So we want to find

Pr[1 = O and (2 = B or 2 = O)]

By independence,

= Pr[1 = O] • Pr[2 = B or 2 = O]

By mutual exclusivity,

= Pr[1 = O] • (Pr[2 = B] + Pr[2 = O])

Then from the given distribution,

= 0.44 • (0.13 + 0.44) = 0.2508 = 25.08%

Now suppose partner 2 is type-B. Then partner 1 must be either type-B or type-O. But the math works out the same way, so that the overall probability is

25.08% + 25.08% = 50.16%

7 0
2 years ago
-29=5(2a-1)+2a i dont. Know how to solve
Aleksandr [31]

Answer:

a = -2 :)))))))))))))))))))))))

7 0
3 years ago
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