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mars1129 [50]
2 years ago
6

HELP ME ASAP!!!!!!!! Simplify these equations

Mathematics
1 answer:
riadik2000 [5.3K]2 years ago
7 0

Step-by-step explanation:

1.3×2×a×b

=6ab

2.c5×c

=c5+1

=c6

3.2y4×5y3

=10y4+3

=10y7

4.3gh2×4g3h3

=12g1+3h2+3

=12g4h5

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Alex73 [517]
Your answer is A. Bro
8 0
3 years ago
Read 2 more answers
Find the average rate of change of the function below over the interval (-4,-1).
babymother [125]

Answer:

-2

Step-by-step explanation:

\frac{f(b)-f(a)}{b-a}\\ \\=\frac{f(-1)-f(-4)}{-1-(-4)}\\ \\=\frac{-4-2}{-1+4}\\ \\=\frac{-6}{3}\\ \\=-2

Therefore, the average rate of change over the interval [-4,-1] is -2.

5 0
2 years ago
Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o
myrzilka [38]

Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that we use the binomial probability distribution 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.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is p = \frac{1}{5} = 0.2

10 questions.

This means that n = 10

A) What is the probability that he will answer all questions correctly?

This is P(X = 10)

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

P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so P(X = 0)

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

P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

P(X \geq 1) = 1 - P(X = 0)

Since P(X = 0) = 0.1074, from item b.

P(X \geq 1) = 1 - 0.1074 = 0.8926

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

In which

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

P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264

P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055

P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008

P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001

P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0

P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0

So

P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

8 0
3 years ago
Jerry borrow $35,00 for 3 years at 7 1/2% simple interest rate how much interest rate he have to pay​
gregori [183]

Answer:

787.50

Step-by-step explanation:

interest=principle*rate*time/100

interest=3500*7.5*3/100

interest=35*7.5*3

interest=787.50

8 0
3 years ago
Simplify the expression
raketka [301]

Answer:

147

Step-by-step explanation:

82+9(12÷3×2)-7

82+9(4×2)-7

82+9(8)-7

82+72-7

154-7

147

Follow PEMDAS

3 0
2 years ago
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