Which graph represents the function f(x) = |x + 3| ?

Elisa withdraw $20 at a time from her bank account and withdraw a total of $140. Ana withdrew $45 at a time from her bank accoun

emmainna [20.7K]

Elisa made a greater number because when you divide $140 by $20 you get 7. So that is seven Twenty dollar bills. For Ana, you divide $270 by $45 and you get 6. So for Ana you get 6 withdrawals of $45.

4 0

3 years ago

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What is 2000000000+ 3000000000

Rom4ik [11]

Answer:

Hope this helps

Step-by-step explanation:

5000000000

4 0

3 years ago

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Write the standard form of the line that passes through the given points. Include your work in your final answer. Type your answ

romanna [79]

Answer:

11x+y = -28

Step-by-step explanation:

We have two points so we can find the slope

m = (y2-y2)/(x2-x1)

= (-6-5)/ (-2--3)

= -11(-2+3)

= -11/1

The slope is -11

We can use point slope to make an equation

y-y1 = m (x-x1)

y-5 = -11(x--3)

y -5 = -11(x+3)

Distribute

y-5 = -11x-33

Add 5 to each side

y-5+5 = -11x-33+5

y = -11x-28

Add 11x to each side

11x+y = -11x+11x-28

11x+y = -28

This is the standard form of a line

5 0

4 years ago

A man invests $5000 in an account that pays 8.5% interest per year, compounded quarterly

notka56 [123]

5000(1+.085/4)^n=1000
to find n use log:
1.02125^n
N X log1.02125
when computed=32.9
round to 33 quarters compounded
33/4
equals to 8 years 3 months

3 0

3 years ago

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The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that

salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

$X \sim Binom(n=20, p=0.411)$

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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

And we can find the individual probability:

$P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253$

And then:

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

And since we want the probability in the 3 classes we can assume independence and we got:

$P= 0.99997^3 = 0.9992$

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

Part a

Let X the random variable of interest, on this case we know that:

$X \sim Binom(n=20, p=0.411)$

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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

And we can find the individual probability:

$P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253$

And then:

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

And since we want the probability in the 3 classes we can assume independence and we got:

$P= 0.99997^3 = 0.9992$

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0

4 years ago