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Anettt [7]
3 years ago
5

Use the number line to find 1/4+-21/8

Mathematics
2 answers:
skad [1K]3 years ago
8 0

Answer:

(Decimal: -2.375) = -19/8

nikdorinn [45]3 years ago
5 0

Answer:

-\frac{19}{8}

Step-by-step explanation:

\frac{1}{4}-\frac{21}{8}=\frac{2}{8}-\frac{21}{8}=\frac{2-21}{8}=-\frac{19}{8}

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The radius of a semicircle is 17 millimeters. what is the semicircle's perimeter?
Maksim231197 [3]

Answer:

87.407

Step-by-step explanation:

circumference = 2*pie*r

= 2*pie*17

=106.8/2

=53.4

53.4+(17*2)

=87.4

4 0
2 years ago
4,864 divsion 2 it multiple 2 word probems
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<h3>4864÷2</h3><h3>2432×2</h3><h3>= 4864</h3>

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3 0
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.

3 0
3 years ago
I'm confused on what to do since I day dream during class, I know it's a bad thing but I have a hard time paying attention​
vagabundo [1.1K]

Answer:17/20 answer a

Step-by-step explanation:

4 0
3 years ago
A line passes through the points (2,-2) and (-6, 2). The point (a, -4) is also one the line. What is the value of a
V125BC [204]

Answer:

b

Step-by-step explanation:

1

3 0
3 years ago
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