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just olya [345]
3 years ago
11

11x - 8y = -4 -4x + 3y = 2

Mathematics
1 answer:
Leto [7]3 years ago
6 0

Answer:

x=4, y=6. (4, 6).

Step-by-step explanation:

11x-8y=-4

-4x+3y=2

----------------

4(11x-8y)=4(-4)

11(-4x+3y)=11(2)

-------------------------

44x-32y=-16

-44x+33y=22

-------------------

y=6

11x-8(6)=-4

11x-48=-4

11x=-4+48

11x=44

x=44/11

x=4

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kirza4 [7]
<h3>Answer:  Choice C) 8.5</h3>

============================================================

Explanation:

Each x represents a data point location.

So, for example, having an x over 60 means 60 is part of the set.

The set of values we're working with is

{59,60,61,63,63,64,66,68,70,71,71,73}

The repeated values are due to the fact we have a stack of two 'x' markers, and they occur at 63 and 71.

To find the IQR (interquartile range), we'll first need to find the median of this set. That's the middle most value.

Count out the number of values to find that there are n = 12 values.

The list splits into two halves that are n/2 = 12/2 = 6 items each

Between slots 6 and 7 is where the median is located.

The value in slot 6 is 64 and the value in slot 7 is 66. Average those two items to get (64+66)/2 = 65

The median is 65

---------------------------------

Next, we'll form two groups L and U such that

L = set of items lower than the median

U = set of items larger than the median

Because n is even, we simply just break the original set into two equal groups (6 items each)

L = {59,60,61,63,63,64}

U = {66,68,70,71,71,73}

The values of Q1 and Q3 represent the medians of L and U in that order.

The median of set L is (61+63)/2 = 62, so Q1 = 62

The median of set U is (70+71)/2 = 70.5, which is Q3

-----------------------------------

To summarize everything so far, we have found

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  • Q3 = 70.5

Subtract those items to get the IQR

IQR = Q3 - Q1

IQR = 70.5 - 62

IQR = 8.5 which points us to choice C as the final answer.

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