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3 years ago

What is slope and y-intercept of the equation on the graph?

Mathematics

1 answer:

san4es73 [151]3 years ago

5 0

The y intercept is the point where the line crosses the y (vertical) axis. So the y intercept is the point (0,3). The slope is 3/2

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Yo, 10⁶ ÷2⅗ simplified?

ziro4ka [17]

10 to the sixth power = 1,000,000

1,000,000 ÷ 2 3/5 = 384615.384615

ANSWER : 384615.384615

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3 years ago

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A cereal company wants to change the shape of its cereal box in order to attract the attention of shoppers. The original cereal

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The new box will both hold more cereal and require more material to make.

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3 years ago

A survey said that 3 out of 5 students enrolled in higher education took at least one online course last fall. Explain your calc

marysya [2.9K]

Answer:

a) 60% probability that student took at least one online course

b) 40% probability that student did not take an online course

c) 12.96% probability that all 4 students selected took online courses.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they took at least one online course last fall, or they did not. The probability of a student taking an online course is independent of other students. 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.

3 out of 5 students enrolled in higher education took at least one online course last fall.

This means that $p = \frac{3}{5} = 0.6$

a) If you were to pick at random 1 student enrolled in higher education, what is the probability that student took at least one online course?

This is P(X = 1) when n = 1. So

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

$P(X = 1) = C_{1,1}.(0.6)^{1}.(0.4)^{0} = 0.6$

60% probability that student took at least one online course.

b) If you were to pick at random 1 student enrolled in higher education, what is the probability that student did not take an online course?

This is P(X = 0) when n = 1.

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

$P(X = 0) = C_{1,1}.(0.6)^{0}.(0.4)^{1} = 0.4$

40% probability that student did not take an online course

c) Now, consider the scenario that you are going to select random select 4 students enrolled in higher education. Find the probability that all 4 students selected took online courses

This is P(X = 4) when n = 4.

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

$P(X = 4) = C_{4,4}.(0.6)^{4}.(0.4)^{0} = 0.1296$

12.96% probability that all 4 students selected took online courses.

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3 years ago

Bugs Bunny charges $13 to baby sit for the first hour, and then charges $3 for every

Klio2033 [76]

Answer:

6 hours

Step-by-step explanation:

both will equal $31 at 6 hours

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3 years ago

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Sin30=cos(2x) then x=

Paraphin [41]

Answer:

x = 30.

Step-by-step explanation:

Use the identity sin A = cos (90 - A).

So sin 30 = cos (90 - 30) = cos 60.

cos 2x = cos 60

so x = 30.

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