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3 years ago

When solving a system of equations, how do you determine which method to use? (substitution, elimination, graphing)

Mathematics

1 answer:

balu736 [363]3 years ago

3 0

It just depends on the equation that you have to solve

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3 years ago

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A multiple choice exam has ten questions. Each question has five possible answers, of which one is correct. Suppose that a stude

Ber [7]

Answer:

8.81% probability that the student answers exactly 4 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 other questions. So 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.

A multiple choice exam has ten questions.

This means that $n = 10$

The probability of answering any question correctly is 0.20.

This means that $p = 0.2$

What is the probability that the student answers exactly 4 questions correctly

This is P(X = 4).

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

$P(X = 4) = C_{10,4}.(0.2)^{4}.(0.8)^{6} = 0.0881$

8.81% probability that the student answers exactly 4 questions correctly

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4 years ago

The fraction 5/8 produces a repeating decimal. ___ 0.625 True or False

Rama09 [41]

Answer:

False i just did it

Step-by-step explanation:

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3 years ago

In how many ways can the letters of the word COMPUTER be arranged in a row if the letters C and O can not stand next to each oth

nirvana33 [79]

Answer:

35280ways

Step-by-step explanation:

let X be the number of ways the letters can be arranged with no restriction,

let M be the number of ways they can be arranged if the letters C and O must stand together,

and let the number of ways the letters C and O can be arranged if they can not stand together.

X=8!

M= 7!

N=X–M=8!–7!=35280 ways

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2 years ago

50% of all the cakes jenny baked that jenny baked last week were party cakes 1/5 were fruit cakes and the remainder were sponge

Juli2301 [7.4K]

30% of them were sponge cakes ( this is because 50% were party cakes and 20% were fruit cakes = 70% which left the sponge cakes, as they were the remainder, to be 30% of what she baked.)

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3 years ago