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bekas [8.4K]
3 years ago
7

What is the median of the following set? {5, 2, 9, 7, 4} A: 5 B: 7 C: 8 D:9

Mathematics
2 answers:
hram777 [196]3 years ago
8 0

A. 5

(2,4,5,7,9)

5 is in the middle between 4 and 7.

Tresset [83]3 years ago
8 0

Answer:

5

Step-by-step explanation:

The median is the number in the middle.

{5, 2, 9, 7, 4}

{2, 4, 5, 7, 9}

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Find the value of c that completes the square for x2−15x+c
meriva

Answer:

The value of <em>c</em> is \frac{225}{4}.

Step-by-step explanation:

The perfect square of the difference between two numbers is:

(x-y)^{2}=x^{2}-2xy+y^{2}

The expression provided is:

x^{2}-15x+c

The expression is a perfect square of the difference between two numbers.

One of the number is <em>x</em> and the other is √<em>c</em>.

Use the above relation to compute the value of <em>c</em> as follows:

x^{2}-15x+c=(x-\sqrt{c})^{2}\\\\x^{2}-15x+c=x^{2}-2\cdot x\cdot\sqrt{c}+c\\\\15x=2\cdot x\cdot\sqrt{c}\\\\15=2\cdot\sqrt{c}\\\\\sqrt{c}=\frac{15}{2}\\\\c=[\frac{15}{2}]^{2}\\\\c=\frac{225}{4}

Thus, the value of <em>c</em> is \frac{225}{4}.

5 0
3 years ago
Simplify this i need help
Temka [501]

Answer:

Step-by-step explanation:

√2x(7 + √2x)

7√2x + √2x ·√2x

7√2x + 2x

or

2x + 7√2x

A is the answer

6 0
3 years ago
Explain how 0 tens and 27 ones and 1 ten and 17 ones show the same answer and the answer is 27
Law Incorporation [45]
Because 0 + 27 is 27 and 10 + 17 is 27 as well 

4 0
3 years ago
Determine the sum of the arithmetic series: 5+18 +31 +44 + ... 161.
Finger [1]

Answer:

1079

Step-by-step explanation:

Hello,

18-5 = 13

31-18=13

44-31=13

161=5+13*12

So we need to compute

\displaystyle \sum_{k=0}^{k=12} \ {(5+13k)}\\\\=\sum_{k=0}^{k=12} \ {(5)} + 13\sum_{k=1}^{k=12} \ {(k)}\\\\=13*5+13*\dfrac{12*13}{2}\\\\=65+13*13*6\\\\=65+1014\\\\=1079

Thanks

3 0
3 years ago
How many elements are in the union of five sets if the sets contain 10,000 elements each, each pair of sets has 1000 common elem
masya89 [10]

Answer:

40951

Step-by-step explanation:

Using the principles of inclusion - Exclusion

Where C(n,r)=n!/(n-r)!r!

Total elements in the five sets including number repetition is given as (10000)×C(5, 1) =10000× 5!/(5-1)!1!=10000×5=50000

Total Number of elements in each pair including number repetition of sets is given as

=(1000) × C(5, 2) =10000

Number of elements in each triple of sets is given as

=(100) × C(5, 3) =1000

Number of elements in every four sets

=(10) × C(5, 4)=50

Number of elements in every one set

(1) × C(5, 5)=1

Therefore total number of unique elements=50000-10000+1000-50+1

=40951

5 0
3 years ago
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