2 years ago

If a single 20 sided dice is tossed once what is the probability of rolling a 15 or 17

1 answer:

6 0

For a single 20 sided dice, a chance to get a certain number would be 1/20. For 2 numbers it will be a 2/20 chance, which can be simplified to a 1/10 chance that you will roll a 15 or a 17.

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I tought of a number double it then added 3. The result multiplied by 4 came to 52,what was the number I tought of?

Digiron [165]

The original number you thought of was 5

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1 year ago

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→ Can someone help me pls ←

Daniel [21]

Where’s the question bro?

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2 years ago

Find lim h->0 f(9+h)-f(9)/h if f(x)=x^4 a. 23 b. -2916 c. 2916 d. 2925

Svetach [21]

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h$

Carry out the binomial expansion in the numerator:

$(9+h)^4 = 9^4+4\times9^3h+6\times9^2h^2+4\times9h^3+h^4$

Then the 9⁴ terms cancel each other, so in the limit we have

$\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h$

Since h is approaching 0, that means h ≠ 0, so we can cancel the common factor of h in both numerator and denominator:

$\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)$

Then when h converges to 0, each remaining term containing h goes to 0, leaving you with

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = 4\times9^3 = \boxed{2916}$

or choice C.

Alternatively, you can recognize the given limit as the derivative of f(x) at x = 9:

$f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h$

We have f(x) = x ⁴, so f '(x) = 4x ³, and evaluating this at x = 9 gives the same result, 2916.

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Final price is $89.40 with the 40% is $59.60

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