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scoray [572]
3 years ago
11

If 160 is divided by x, then the remainder is 4. What is the remainder when 315 is divided by x?

Mathematics
1 answer:
frutty [35]3 years ago
4 0
Let us rewrite this equation as:
<em>160/x = some number + 4/x
</em>we can assume that some number of x's can fit into 160, but x is greater than 4 so we have a remainder.
We can figure out by using the remainder that x is completely divisible into (160-4) or 156. Knowing this we can assume that x is some factor of 156 that is greater than 4. 
Let us list the factors of 156
1,2,3,4,6,12,13,26,39,52,78,156
Now let us refer back to the restrictions we put on x.
<em>x must be > 4 as anything less than 4 would divide at least 1 more time into it. So now we are left with a bunch of other factors. And the cool thing is that all of the factors excluding anything that is = or < 4 all work for getting a remainder of 4. Thus we can use the largest factor and set it equal to x.
</em>So x = 156
then if we divide 315 by x or <em>315/x = 315/156 = 2 and a remainder of 3
so the answer would be 3.</em>
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Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
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Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

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a) Toss 2 coins; record the order of heads and tails.

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When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

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3 years ago
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