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Gala2k [10]
3 years ago
11

Which two points satisfy y = -x2 + 2x + 4 and x + y = 4?

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
4 0
y=-x^2+2x+4 
x+y=4
y=4-x
4-x=-x^2+2x+4
x^2-3x=0
x(x-3)=0
x=0 or x =3

for x = 0 y = 4
for x = 3 y = 1

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f(x)=8−4x−x3 g(x)=x2+7x−9 Find f(x)+g(x). Select one: a. x3+x2+3x−1 b. −x3+x+3x−1 c. −x3+x2+11x−9 d. 8x2+3x−9x3
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We are given the two functions:

f(x) = 8 - 4x - x^3\text{ and } g(x) = x^2 + 7x - 9

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= (8 - 4x - x^3) + (x^2 + 7x - 9)

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Step-by-step explanation:

The steps to find an outlier:

1. Put the data in numerical order.

2. Find the median.

3. Find the medians for the top and bottom parts of the data. This divides the data into 4 equal parts.

The median with the smallest value is called Q1. The median for all the values - usually just called the median is also called Q2. The median with the largest value is Q3.

4. Subtract...Q3 - Q1. This value is the InterQuartileRange or IQR. Remember that the range means taking the largest minus the smallest. This is a special range having to do with the quartiles.

5. Multiply...1.5 * IQR

6. Take your answer from #5 and do 2 things with it. A). Subtract it from Q1 and B) Additional to Q3.

7. Look at all your data points. If any are SMALLER than Q1 - 1.5 *IQR, they are outliers. If any are LARGER than Q3 + 1.5 *IQR, they are also outliers.

For your data....the median, Q2 is

(43+38)/2 = 40.5.

Q1 = (30+26)/2 = 28.

Q3 = (54+52)/2 = 53

The IQR is 53 - 28 = 25

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Q1 - 37.5 = 28 - 37.5 = -9.5. There is no data value less than -9.5.

Q3 + 37.5 = 53 + 37.5 = 90.5. there is no data value greater than 90.5.

My conclusion is that there are no outliers in this data.

I hope this helps!

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