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2 years ago

Help help help help help help please

Mathematics

2 answers:

ololo11 [35]2 years ago

5 0

Hello there :)

The first option: a^m-n

This is the rule quotient of powers

Good luck, u r amazing

joja [24]2 years ago

3 0

${a}^{m} \div {a}^{n} = {a}^{m - n}$

<h3>Explanation :</h3>

${a}^{m} \div {a}^{n} = {a}^{m - n}$

${a}^{m} \times {a}^{n} = {a}^{m + n}$

$( {a}^{m}){}^{n} = {a}^{m \times n}$

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Can some break this down step by step please I'm confused?

frutty [35]

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3 years ago

What is the value of f(x) when x =0

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Answer:

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Step-by-step explanation:

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2 years ago

The equation 2 x - 4 y = 7 is written in slope-intercept form. True False

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2 years ago

A starter motor used in a space vehicle has a high rate of reliability and was reputed to start on any given occasion with proba

GuDViN [60]

Answer:

0.095163

Step-by-step explanation:

given that a starter motor used in a space vehicle has a high rate of reliability and was reputed to start on any given occasion with probability .99999

Here we find that for any start, there are exactly two outcomes either success or failure.

Also each start is independent of the other since p = 0.99999 for succss is given constant.

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If Y is taken as failure then Y is binomial with p' = 0.00001 and n =10000

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3 years ago

There are 14 juniors and 16 seniors in a chess club. a) From the 30 members, how many ways are there to arrange 5 members of the

Dmitrij [34]

Answer:

a) 17,100,720

b) 4,717,440

c) 10,920

d) 2821

Step-by-step explanation:

14 juniors and 16 seniors = 30 people

a) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

As it is a ordered arrangement

30.29.28.27.26 = 17,100,720

b) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

16.28.27.26.15 = 4,717,440

c) If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Not ordered arrangement. And means that we need to multiply the results.

C₁₄,₂ * C₁₆,₂

C₁₄,₂ = 14.13.12! = 14.13 = 91

12! 2! 2

C₁₆,₂ = 16.15.14! = 16.15 = 120

14! 2! 2

C₁₄,₂ * C₁₆,₂ = 91.120 = 10,920

d) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

Or means that we need to sum the results.

C₁₄,₄ + C₁₆,₄

C₁₄,₄ = 14.13.12.11.10! = 14.13.12.11 = 1001

10! 4! 4.3.2.1

C₁₆,₄ = 16.15.14.13.12! = 16.15.14.13 = 1820

12! 4! 4.3.2.1

C₁₄,₄ + C₁₆,₄ = 1001 + 1820 = 2821

7 0

3 years ago

Help help help help help help please​

Help help help help help help please