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4 years ago

Which equation is in standard form?

Mathematics

2 answers:

Sauron [17]4 years ago

6 0

Standard form is

ax + by = c

Only choice A is in that form.

levacccp [35]4 years ago

3 0

Answer: C

Step-by-step explanation: took the same assignment, I might be wrong but hey, I'm not the only one answering

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3.<br> Find two numbers whose product is -6 and whose sum is 1.

Anestetic [448]

Answer:

-2, 3

Step-by-step explanation:

yeah that's it I don't have any explanation or formula

5 0

3 years ago

Read 2 more answers

Travelers who fail to cancel their hotel reservations when they have no intention of showing up are commonly referred to as no-s

notsponge [240]

Answer:

a) 0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) 0 is the most likely value for X.

Step-by-step explanation:

For each traveler who made a reservation, there are only two possible outcomes. Either they show up, or they do not. The probability of a traveler showing up is independent of other travelers. 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.

No-show rate of 10%.

This means that $p = 0.1$

Four travelers who have made hotel reservations in this study.

This means that $n = 4$

a) What is the probability that at least two of the four selected will turn to be no-shows?

This is $P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)$

In which

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

$P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486$

$P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036$

$P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001$

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0486 + 0.0036 + 0.0001 = 0.0523$

0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) What is the most likely value for X?

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

$P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.6561$

$P(X = 1) = C_{4,1}.(0.1)^{1}.(0.9)^{3} = 0.2916$

$P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486$

$P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036$

$P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001$

X = 0 has the highest probability, which means that 0 is the most likely value for X.

7 0

3 years ago

how do you solve -2.8k-4.31=-15.79 or 1.9m-3=-12.31 or 2.1a+1.4=-14.245 or 3.8n-4.8=-21.9 ?

Effectus [21]

Consider posting each question separately, please.

Given <span>-2.8k-4.31=-15.79 :

1) Isolate the "k" term on the left by adding 4.31 to both sides of this equation.
2) Simplify the right side.
3) Divide both sides by -2.8 and you'll have the answer: k = ?</span>

5 0

3 years ago

Roger bring 4 cakes for desert to the community potluck feast and powwow. His son brings 2 individual servings of dessert. Altog

TiliK225 [7]