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Prove the following using a proof by contradiction:The average of four real numbers is greater than or equal to at least one of

skelet666 [1.2K]

Answer:

By contradiction it can be proved that:

Average of four real numbers will be greater than or equal to at least one of the four real numbers.

Step-by-step explanation:

Method of contradiction means we assume opposite to the facts to be proved and then we contradict our assumption.

As a result, we prove the fact.

Here, let the four number be:

$p, q, r, s$

The average will be Sum of all numbers divided by count of numbers.

$\dfrac{p+ q+ r+ s}4$

Now, let us assume the opposite that the average is less than all the numbers.

i.e.

$\dfrac{p+ q+ r+ s}4$

Now, let us add all of them:

$\dfrac{p+ q+ r+ s}4 +\dfrac{p+ q+ r+ s}4 +\dfrac{p+ q+ r+ s}4 +\dfrac{p+ q+ r+ s}4 < p+q+r+s\\\Rightarrow \dfrac{4(p+q+r+s)}4$

Which can never be possible hence, our assumption is contradicted.

our assumption is wrong.

Therefore, by contradiction it is proved that:

Average of four real numbers will be greater than or equal to at least one of the four real numbers.

8 0

2 years ago

Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x, y) = xe−x(1 + y) x ≥ 0 an

Yuri [45]

Answer:

For the given explanation we see that two life times are not independent

Step-by-step explanation:

probability for X (for x≥ 0)

$\int\limits^\infty_0 {xe^-^x^(^1^+^y^)} \, dy$

$-e^-^x\int\limits^\infty_0 {de^-^x^y} \,$

e⁻ˣ

Probability for X exceed 3

= $\int\limits^\infty_3 {f(x)dx$

= $\int\limits^\infty_3 {e^-^3 dx$

= $e^-^3$

probabilty for y≥ 0 is

$\int\limits^\infty_0 {xe^-^x^(^1^+^y^)} \, dx \\$

$\int\limits^\infty_0{-x/1+yd(e^-^x^(^1^+^y^))} \, =1/(1+y)² \\$

3 0

2 years ago

An ostrich can run 6 mph faster than a giraffe. An ostrich can run 7 miles in the same time that a giraffe can run 6 miles. Find

LUCKY_DIMON [66]

NOTE THIS IS AN EXAMPLE:

Let t = time, s = ostrich, and g = giraffe.

Here's what we know:

s = g + 5 (an ostrich is 5 mph faster than a giraffe)

st = 7 (in a certain amount of time, an ostrich runs 7 miles)

gt = 6 (in the same time, a giraffe runs 6 miles)

We have a value for s, so plug it into the first equation:

(g + 5)t = 7

gt = 6

Isolate g so that we can plug that variable value into the equation:

g = 6/t

so that gives us:

(6/t + 5)t = 7

Distribute:

6 + 5t = 7

Subtract 6:

5t = 1

Divide by 5:

t = 1/5 of an hour (or 12 minutes)

Now that we have a value for time, we can plug them into our equations:

1/5 g = 6

multiply by 5:

g = 30 mph

s = 30 + 5

s = 35 mph

Check by imputing into the second equation:

st = 7

35 * 1/5 = 7

7 = 7

5 0

2 years ago

What is the horizontal asymptote of f (x) =StartFraction negative 2 x Over x + 1 EndFraction? y = –2 y = –1 y = 0 y = 1

miv72 [106K]

Answer:

y = -2

Step-by-step explanation:

Any asymptotes of a rational function will be described by the quotient of the numerator and denominator (excluding any remainder).

$f(x)=\dfrac{-2x}{x+1}=-2+\dfrac{2}{x+1}$

The horizontal asymptote is ...

y = -2

8 0

2 years ago

Read 2 more answers

A cement bridge post is 24 inches square and 15 feet 9 inches in length. If the cement weighs 145 pounds per cubic foot, how muc

lisov135 [29]

Answer:

The answer is option (C), One cement bridge post weighs 9,135 pounds

Step-by-step explanation:

Step 1: Get the total volume of a cement bridge post

The cement bridge post is in the shape of a cuboid, therefor the volume of the cement bridge post can be expressed as;

Volume of cement bridge post=Base area×Length

where;

Base area=(24^2) inches square

1 foot=12 inches

Convert 24 inches to foot=24/12=2^2=4 feet²

Length=15 feet and 9 inches

1 foot=12 inches

Convert 9 inches to foot=9/12=0.75 feet

Total length=(15+0.75)=15.75 feet

replacing;

Volume of cement bridge post=(4×15.75)=63 cubic feet

Volume of cement bridge post=63 cubic feet

Step 2: Get the total weight of the cement bridge post

Total weight of the cement bridge post=Weight per cubic foot×total volume of the cement bridge post

where;

Weight per cubic foot=145 pounds per cubic foot

Total volume of the cement bridge post=63 cubic feet

replacing;

Total weight of the cement bridge post=(145×63)=9,135 pounds

One cement bridge post weighs 9,135 pounds

8 0

2 years ago