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FinnZ [79.3K]
2 years ago
5

Please Help!! What should be done 10x ^ 2 - 8x to create a perfect square ?

Mathematics
1 answer:
Pani-rosa [81]2 years ago
4 0

Answer: 10x^2-8x+16

Step-by-step explanation:

Use the formula (b/2)^2

then plug in values:

(-8/2)^2 = -4^2 = 16

ans: 10x^2-8x+16

hope this helps

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What fraction is bigger 5/8 or 7/15
Zolol [24]
5/8=.625 and 7/15=.466 repeating so 5/8 is a bigger fraction
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3 years ago
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A network provider investigates the load of its network. The number of concurrent users is recorded at fifty locations (thousand
beks73 [17]

Answer:

see explaination

Step-by-step explanation:

Using the formulla that

sum of terms number of terms sample mean -

Gives the sample mean as \mu=17.954

Now varaince is given by

s^2=\frac{1}{50-1}\sum_{i=1}^{49}(x_i-19.954)^2=9.97

and the standard deviation is s=\sqrt{9.97}=3.16

b) The standard error is given by

\frac{s}{\sqrt{n-1}}=\frac{3.16}{\sqrt{49}}=0.45

c) For the given data we have the least number in the sample is 12.0 and the greatest number in the sample is 24.1

Q_1=15.83, \mathrm{Median}=17.55 and Q_3=19.88

d) Since the interquartile range is Q_3-Q_1=19.88-15.83=4.05

Now the outlier is a number which is greater than 19.88+1.5(4.05)=25.96

or a number which is less than 15.83-1.5(4.05)=9.76

As there is no such number so the given sample has no outliers

5 0
3 years ago
Compute the permutations and combinations. How many different arrangements can be made from the letters of the word "MISSISSIPPI
Eddi Din [679]
The answer is the last one

I hope that helped
4 0
3 years ago
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What number should be added to both sides of the equation to complete the square? x2 – 6x = 5
choli [55]

Answer: Add 9 to both sides to complete the square

Step-by-step explanation:

The formula for the number you need to add to complete the square is (b/2)^2. In this case, b = -6, so:

(b/2)^2 = (-6/2)^2 = (-3)^2 = 9.

(After adding 9 to both sides, you will get x^2 - 6x + 9 = 14 which can be factored to (x - 3)^2 = 14)

3 0
3 years ago
A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1
KATRIN_1 [288]

Answer:

Step-by-step explanation:

Heres the complete question:

A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1990–94, and 1995–99. If the population at the end of 1999 was 9,320:

How many people lived in the town at the beginning of 1985? (Round your answer to the nearest whole number.)

solution:

Let the population of the town at the beginning of 1985 be P. Then, given that in the first five-year period the population declined by 3.2%, i.e., 0.032, the population of the town at the end of 1989 would be

(1 – 0.032)P = 0.968P.

Again, given that in the second five-year period the population declined by 5.2%, i.e., 0.052, the population of the town at the end of 1994 would be

(1 – 0.052)(0.968P) = 0.948 x 0.968P = 0.917664P.

Finally, given that in the third five-year period the population declined by 4.7%, i.e., 0.047, the population of the town at the end of 1999 would be

(1 – 0.047)(0.917664P) = 0.874533792P.

We are given, 0.874533792P = 9320 or

P = 9320/0.874533792 = 10657.11.

Thus, 10657 people lived in the town at the beginning of 1985

3 0
3 years ago
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