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Sholpan [36]
3 years ago
9

An astronaut who weighs 174 lb on Earth weighs 29 lb on the moon. Which of the following are correct amounts for weight on Earth

compared to weight on the moon?
A.102 lb and 17 lb

B.66 lb and 12 lb

C.170 lb and 34 lb

D.105 lb and 15 lb
Mathematics
2 answers:
Korolek [52]3 years ago
5 0

174/29 = 6 times the difference

17*6 = 102


12*6= 72

34*6 =204

15*6 = 90


so the answer is A

timofeeve [1]3 years ago
3 0

The answer  to this question is A

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To solve this you must use a proportion like so...

\frac{part}{whole} = \frac{part}{whole}

The total number of students that can be chosen are 4,663. This number will represent the whole of one fraction in the proportion. We want to know what percent probability out of these students are engineer, medical doctor/surgeon. This would be considered the part of this fraction. Sum the number of engineering students (615) with medical doctors/surgeons (723) to find this number

723 + 615 = 1,338 students that want to be an engineer or medical doctor/surgeon

Percent's are always taken out of the 100. This means that the other fraction in the proportion will have 100 as the whole and x (the unknown) as the part.

Here is your proportion:

\frac{1,338}{4,663} =\frac{x}{100}

Now you must cross multiply

1,338*100 = 4,663*x

133,800 = 4,663x

To isolate x divide 4,663 to both sides

133,800/4,663 = 4,663x/4,663

28.7 = x

This means that there is a 28.7% of a student with the intent of becoming an engineer or a medical doctor/surgeon to be chosen at random

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Answer:

The expectation is  E(1 )= -\$ 1

Step-by-step explanation:

From the question we are told that  

     The first offer is  x_1 =  \$ 8000

     The second offer is  x_2 =  \$ 4000

      The third offer is  \$ 1600

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      The  price of each ticket is  p= \$ 5

Generally expectation is mathematically represented as

             E(x)=\sum  x *  P(X = x )

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    P(X =  x_1  ) = 0.0002    

 Now  

     P(X =  x_2  ) =  \frac{1}{5000}    given that they just offer one

     P(X =  x_2  ) = 0.0002    

 Now  

      P(X =  x_3  ) =  \frac{5}{5000}    given that they offer five

       P(X =  x_3  ) = 0.001

Hence the  expectation is evaluated as

       E(x)=8000 *  0.0002 + 4000 *  0.0002 + 1600 * 0.001

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Now given that the price for a ticket is  \$ 5

The actual expectation when price of ticket has been removed is

      E(1 )= 4- 5

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Answer:

the first statement ( I )

Step-by-step explanation:

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Answer:

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Skip,Flip,Multiply Method

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