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harina [27]
2 years ago
11

2,114 x 98 *100 points show work

Mathematics
1 answer:
Vlad [161]2 years ago
6 0
I can show you on a piece of paper
Hope this helps *smiles*

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Reduce 36/45 to lowest terms and write the numerator in the blank.
saveliy_v [14]
4/5 is your answer

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Which is the solution of this system of equations? y=5x−83y=x+18<br><br> y+5(3)=15
Ulleksa [173]

Answer:

y=5x−83y=x+18    (the question is wrong; it has 2 equal signs)

y+5(3)=15

y+15=15   (multiply 5*3; you get 15)

y= 15-15

y= 0

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The triangles shown are congruent which of the following must be true
Fantom [35]
The bottom answer m y = m h
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The probability that a person in the United States has type B​+ blood is 10%. Four un-related people in the United States are se
marysya [2.9K]

Answer:

a. 0.0001

b. 0.6561

c. 0.3439

d. B. The event in part​ (a) is unusual because its probability is less than or equal to 0.05.

Step-by-step explanation:

a. # We are given that the probability that a person in the United States has Type B+ blood = 0.10. Also we are told that four unrelated people in the United States are selected at random.

#We have to find here the probability that all four have type B+ blood.

Since the events are independent, we have :

Probability that all four have B+ blood  = 0.10 x 0.10x 0.10x0.10

                                                                                       = 0.0001

Therefore, the probability that all four have type B+ blood is 0.0001

b. We have to find the probability that none have B+ blood. Using the complementary law of probability we have:

Probability that blood type is not B+ = 1 - 0.10= 0.90                                                                        

Therefore, the probability that none have B+ blood

= 0.90 x 0.90 x 0.90x0.90=0.6561

Therefore, the probability that none have B+ blood is 0.6561

c. We have to find the probability that at least one of the four have B+ blood.

#The probability that at least one of the four have B+ blood = 1 -  Probability that none have B+ blood type

=1-0.6561=0.3439

Therefore,the probability that at least one of the four has type B+ blood is 0.3439

d. An event is considered unusual if the probability of the event is small or less than 0.05 . We note that event a is the only small probabilty and is less than 0.05.

-a is thus considered unusual(the rest are all usual events)

                                                                                                                 

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2 years ago
Subtract express each diffrence in simplest form : 2 - 1/3-9/10 and 4-5/6- 3/8
ivolga24 [154]
The first one is 23/30  and the second one is 2 19/24
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