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3 years ago

A(-1, 3), B(2, 2), and C(3, -6). If the triangle is translated 5 units to the right, what are the coordinates of A'?

Mathematics

1 answer:

Jobisdone [24]3 years ago

3 0

Answer:

(4,3)

Step-by-step explanation:

If the triangle ABC is moved 5 units to the right, the coordinates of A will be (4,3). We can find this by using the values already given for A. We are given (-1,3) which is A's current position. Since we are moving it to the right we are only looking at the first number. If we add 5 to -1 we get 4 which we put in the place of the -1 in the coordinates to get our answer.

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In a particular faculty 60% of students are men and 40% are women. In a random sample of 50 students what is the probability tha

zimovet [89]

Answer:

a) The expected value is given by:

$E(X) = np = 50*0.4 = 20$

and the variance is given by:

$Var(X) =np(1-p) = 50*0.4*(1-0.4) = 12$

b) $P(X>25)= 1-P(X\leq 25)$

And we can find this probability with the following Excel code:

=1-BINOM.DIST(25,50,0.4,TRUE)

And we got:

$P(X>25)= 1-P(X\leq 25)=0.0573$

c) 1) Random sample (assumed)

2) np= 50*0.4= 20 >10

n(1-p) =50*0.6= 30>10

3) Independence (assumed)

Since the 3 conditions are satisfied we can use the normal approximation:

$X \sim N(\mu = 20 , \sigma= 3.464)$

d) $P(X>25) = 1-P(Z< \frac{25-20}{3.464}) = 1-P(z$

e) $P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)$

$P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)= 1-P(Z$

Step-by-step explanation:

Previous concepts

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".

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

$X \sim Binom(n=50, p=0.4)$

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:

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

Part a

The expected value is given by:

$E(X) = np = 50*0.4 = 20$

and the variance is given by:

$Var(X) =np(1-p) = 50*0.4*(1-0.4) = 12$

Part b

For this case we want to find this probability:

$P(X>25)= 1-P(X\leq 25)$

And we can find this probability with the following Excel code:

=1-BINOM.DIST(25,50,0.4,TRUE)

And we got:

$P(X>25)= 1-P(X\leq 25)=0.0573$

Part c

1) Random sample (assumed)

2) np= 50*0.4= 20 >10

n(1-p) =50*0.6= 30>10

3) Independence (assumed)

Since the 3 conditions are satisfied we can use the normal approximation:

$X \sim N(\mu = 20 , \sigma= 3.464)$

Part d

We want this probability:

$P(X>25) = 1-P(Z< \frac{25-20}{3.464}) = 1-P(z$

Part e

For this case we use the continuity correction and we have this:

$P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)$

$P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)= 1-P(Z$

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