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3 years ago

Of the following data, 17, 20,14,5,30,13,6, and 5 find the mean median, and mode

Mathematics

1 answer:

Kazeer [188]3 years ago

6 0

Arrange in ascending order:-

5 5 6 13 14 17 20 30

Mode = 5

Median = mean of middle numbers = (13+14) / 2 = 13.5

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Icy River increased the price of a $1.15 cola by 15%. Find the increase in price and the new price. round to the nearest cent

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15% x
100 1.15 X is your variable, your unknown number and 15% has to always be over 100 , Then you have to cross multiply and divide 100 because that is the only number basically you have to just multiply across and divide 100 to make it to the nearest cent you have to round the last two numbers after the decimal Representing money.. Hope this helps!!

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3 years ago

If you bought a stock last year for a price of $139, and it has risen 5.4% since

kumpel [21]

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Step-by-step explanation:

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3 years ago

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3 years ago

Suppose that f : [a, b] → [a, b] is continuous. Prove that f has a fixed point. That is, prove that there exists c ∈ [a, b] such

nikklg [1K]

Answer:

Step-by-step explanation:

define the function:

$g(x) = f(x) -x$

As both $f(x)$ and x are continuous functions, $g(x)$ will also be continuous.

Now, what can we say about $g(a) = f(a) -a$ ?

we know that $a\leq f(a) \leq b$ , thus:

$a-a\leq f(a)-a \leq b-a\\0 \leq g(a) \leq b-a$

thus $g(a)$ is non-negative.

What about $g(b)$ ? Again we have:

$a\leq f(b) \leq b\\a-b \leq f(b) -b \leq 0\\a-b \leq g(b) \leq 0$

That means that $g(b)$ is not positive.

Now, we can imagine two cases, either one of $g(a)$ or $g(b)$ is equal to zero, or none of them is. If either of them is equal to zero, we have found a fixed point! In fact, any point $c$ for which $g(c)=0$ is a fixed point, because:

$g(c) = 0 \implies f(c) -c = 0 \implies f(c) = c$

Now, if $g(a) \neq 0$ and $g(b) \neq 0$ , then we have that

$g(a) >0$ and $g(b) < 0$ . And by Bolzano's theorem we can assert that there must exist a point c between a and b for which $g(c)=0$ . And as we have shown before that point would be a fixed point. This completes the proof.