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3 years ago

PLEASE HELP ASAP

Mathematics

2 answers:

Vlad1618 [11]3 years ago

6 0

Answer:

Step-by-step explanation:

it easy

lbvjy [14]3 years ago

3 0

Answer:

give the range of the function?

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HELP ASAP PLEASE

yulyashka [42]

A is your answer. Parallel lines never intersect. They both run the same direction, the same distance from each other.

_________________________

_________________________

are parallel lines

7 0

3 years ago

Add 212+(−13) . Write your answer as a mixed number in simplest form.

nordsb [41]

The value of the given expression after addition is 199 and this can not written as a mixed number in the simplest form.

According to the question,

We have the following expression:

212+(-13)

Now, we know that the multiplication of one negative and one positive integers is always positive.

So, we have the following expression:

212-13

199

Now, this number is whole number and it can not be written as a mixed number. It could only have been written as a mixed number when the result would have been in a fraction.

Hence, the value of the given expression after addition is 199 and this can not written as a mixed number in the simplest form.

To know more about addition here

brainly.com/question/19867930

#SPJ1

7 0

1 year ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

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".

Solution to the problem

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

$X \sim Binom(n=20, p=0.2)$

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

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago

An urn contains six red balls and four green balls. A sample of seven balls is selected at random.

KIM [24]

There are 10 balls in the Urn Total.

Red: 6

Green: 4

Question One: The probability that five red and two green is selected is likely. (as that is over half for both)

Question Two: Impossible. There is only 6 red balls, and 7 are taken from the urn. Thus it would at most be possible for 6 red and 1 green.

Question Three: At least four is likely, as there is more red then green in the Urn.

Hope I helped!

(Mark Brainliest if you can please!)

5 0

3 years ago

PLEASE ANSWER FAST IM FAILING MATH!!!

posledela

Answer:

Step-by-step explanation:

From the figure attached,

Point B has been dilated to form point B'.

B(3, 1) → B'(6, 2)

→ B'[(2 × 3), (2 × 1)]

Since rule for the dilation of a point (x, y) by a factor of k is,

B(x, y) → B'(kx, ky)

By comparing the coordinates k = 2 is the scale factor by which the point B has been dilated about the origin.

Therefore, other vertices of the quadrilateral will be,

A(-2, 3) → A'(-4, 6)

C(1, -1) → C'(2, -2)

D(-3, -2) → D'(-6, -4)

4 0

3 years ago