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3 years ago

Given x less than y, compare the following expressions and determine which is greater: 2x-y;2y-x. Explain your answer.

1 answer:

8 0

Well in this case, to see which is greater, let's just say that:
x=4 <--- 4 is less than 5
y=5

Now substitute:
2x-y 2y-x
2(4)-5 2(5)-4
8-5 10-4
3 6

Which of these is greater, 3 or 6? 6, obviously. Therefore, we know that 2y-x is greater.
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Nancy shot a 16 on 4 holes of golf. At this rate, what can she expect her score to be if she plays 18 hole? Round to the nearest

Nataliya [291]

Answer:

108.

Step-by-step explanation:

"At this rate, ..." means we're to assume Nancy's score is proportional to the number of holes:

score/18 = 24/4

Multiplying by 18, we get ...

score = 18(24/4) = 18(6) = 108

Nancy can expect a score of 108 on 18 holes.

(if this isn't right, tell me and I will make sure to recheck my work :).

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2 years ago

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I need help with this story problem.. 8th Grade Chapter 9.3 Linear Functions:

Paladinen [302]

(1/5)x + 8 This will find 20% of the wholesale cost and add the $8 markup, where 'x' is the wholesale cost

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3 years ago

Please help! - Kim needs to place 32 mystery books and 56 biography books on library shelves. She is to place the maximum number

Oksi-84 [34.3K]

A. 8 books on each shelf
B. 4 shelves for mystery, 7 shelves for biography

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3 years ago

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Mike had 119 dollars to spend on 6 books. after buying them he had 11 dollars. how much did each book cost ?

deff fn [24]

18 $ Hope I helped
This is how I came to the result..
119-11= 108
108 : 6 = 18 $

8 0

2 years ago

Consider rolling two fair dice one 3-sided the other 5-sided

Ne4ueva [31]

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

$1\leq x\leq 3,\quad 1\leq y \leq 5$

We have $P(X=x)=\frac{1}{3}$ and $P(Y=y)=\frac{1}{5}$ , because the dice are fair.

Now we use the assumption of independence to claim that

$P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}$

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

2 in a unique way (1+1)
3 in two possible ways (1+2, 2+1)
4 in three possible ways
5 in three possible ways
6 in three possible ways
7 in two possible ways
8 in a unique way

This implies that the probabilities of the outcomes of $W=X+Y$ are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5

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2 years ago