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marusya05 [52]
3 years ago
7

Simplify a⁵ × (a³)²​

Mathematics
1 answer:
FrozenT [24]3 years ago
3 0

Answer:

a^{11}

Step-by-step explanation:

In solving for a⁵ × (a³)², we need to follow the rules of PEMDAS

Therefore, your equation would look like this:

a⁵ × (a³)² = a⁵ × a^{6\\

When two variables are the same, we add their exponents when multiplying and we would subtract them if we divided.

With that information, your final answer would be

a^{11} as 5+6 = 11

Hope this Helps!

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Answer:  The required probability is 0.26.

Step-by-step explanation: Given that there are 60 red marbles and 40 blue marbles in a  box 10 marbles are picked without replacement.

We are to find the probability of selecting 6 red marbles.

Since the marbles are picked up without replacement, so it is a situation of combination.

Let S denote the sample space of the experiment of drawing 10 marbles and E denote the event that 6 marbles are red.

So,

n(S)=^{100}C_{10}=\dfrac{100!}{10!(100-10)!}=\dfrac{100!}{10!90!}=17310309456440,\\\\\\n(E)=^{60}C_6\times^{40}C_4=\dfrac{60!}{6!54!}\times\dfrac{40!}{4!36!}=50063860\times91390.

Therefore, the probability of event E is given by

P(E)=\dfrac{n(E)}{n(S)}=\dfrac{50063860\times91390}{17310309456440}=0.26.

Thus, the required probability is 0.26.

6 0
3 years ago
(5h​3​​−8h)+(−2h​3​​−h​2​​−2h)
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<span>(5h^​3 ​​− 8h) + (−2h​^3 ​− h^​2 ​​− 2h)
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3 years ago
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7 0
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Assume that there are an equal number of births in each month so that the probability is that a person chosen at random was born
NikAS [45]

Answer:

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have a birthday in May, or they do not. The probability of a person having a birthday in May is independent of any other person. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability of a person being in May:

May has 31 days in a year of 365. So

p = \frac{31}{365} = 0.0849

Group of 20 friends:

This means that n = 20

What is the probability that at the May celebration, exactly two members of the group have May birthdays?

This is P(X = 2).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{20,2}.(0.0849)^{2}.(0.9151)^{18} = 0.2773

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

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3 years ago
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Answer:

4/8

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