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Nat2105 [25]
3 years ago
9

You are going to draw two candy bars from a large bag containing 28 snickers bars and 5 twix bars. what is the probability that

both candy bars you draw from the bag are snickers?
Mathematics
1 answer:
amm18123 years ago
6 0
Find the probability of drawing 1 snicker first then the probability of drawing another snicker after that, then multiply them together. For the first choice, the probability of drawing a Snickers is 28/33 (snickers/total candy bars). For the second choice, there are only 32 candy bars left, 27 of which are Snickers. So that probability is 27/32, so...
The probability of BOTH things happening is their product 756/1,056 which is 63:88
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Huh...? You said free points...

Step-by-step explanation:

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Tell whether b=82 is a solution of b÷9=9
netineya [11]

Answer:

no, it's 81

Step-by-step explanation:

9×9=81 hope that this helps you and have a great day :)

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Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1
belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

            F(x) = \int\limits^a_b {f(x)} \, dx

By using integration formula

\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

 \int\limits^6_^-6} (x^{5}  + 8 x^{4}  + 2 x^{2}  + 5 x + 15) )dx

<em>On integration , we get</em>

=   (\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}

F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)

= (\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))

After simplification and cancellation we get

 =  \frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6

on calculation , we get

= \frac{124,416}{5} + \frac{864}{3} + 180

On L.C.M  15

= \frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}

= 25 351.2 units²

<u><em>Conclusion</em></u>:-

<em>The area of the region between the graph of the given function and the x-axis = 25,351 units²</em>

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