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Oduvanchick [21]
2 years ago
6

How do I even figure this out??

Mathematics
1 answer:
kiruha [24]2 years ago
5 0

The value of a is 1/64.

The rules of the exponent are

(a) a^m*a^n=a^{m+n

(b)a^m/a^n=a^{m-n

(c)(a^b)^c=a^{bc}

(d) \sqrt[n]{a}=a^{1/n

(e) a⁻ⁿ=1/aⁿ

From the table it is clear that

By the rule of the exponent a⁻ⁿ=1/aⁿ

where a=2

2⁻¹ = 1/2¹ = 1/2

2⁻² =1/2² =1/4

2⁻³ =1/2³ =1/8

2⁻⁴ =1/2⁴ =1/16

2⁻⁵ =1/2⁵ =1/32

So we have to calculate the left one where it is given that we have to estimate the value for 2⁻⁶ which is a.

From the above, it is clear that 2⁻ⁿ =1/2ⁿ

using the above formula

2⁻⁶= 1/2⁶ =1/64= a

Therefore the value of a is 1/64.

Learn more about the exponent

here: brainly.com/question/535578

#SPJ10

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laiz [17]

<u>Answer-</u>

a. Probability that  three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

<u>Solution-</u>

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

\binom{19}{3} = \frac{19!}{3!16!} = 969

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

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\binom{33}{6} = \frac{33!}{6!27!} =1107568

So total number of selection = (969)×(1107568) = 1073233392

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\binom{52}{9} = \frac{52!}{9!43!} = 3679075400

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Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,

\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}

=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})

=(969)(45)(7)(5)(15)=22892625

Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,

\binom{52}{9}=\frac{52!}{9!43!} =3679075400

∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =

\frac{22892625}{3679075400} = 0.006


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