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3 years ago

Y=x-1 and y=-x+7 how do I graph it

1 answer:

7 0

Find the value of x using substitution

(x-1)=-x+7
2x-1=7
2x=6
x=4

plug your x value into one of the equations to find y

y=(4)-1
y=3

your coordinate is (4,3) you would plot that point.

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What's absolute value |-8|

maks197457 [2]

The absolute value of a number can never be negative. That's because it represents the distance of a number from zero on a number line. The number $-8$ is $8$ digits to the left of zero on a number line. So, the absolute value of $-8$ would be $8$ .

Let me know if you have any questions regarding this problem!
Thanks!
-TetraFish

7 0

4 years ago

Read 2 more answers

Travelers who fail to cancel their hotel reservations when they have no intention of showing up are commonly referred to as no-s

notsponge [240]

Answer:

a) 0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) 0 is the most likely value for X.

Step-by-step explanation:

For each traveler who made a reservation, there are only two possible outcomes. Either they show up, or they do not. The probability of a traveler showing up is independent of other travelers. This means that the binomial probability distribution is used 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.

No-show rate of 10%.

This means that $p = 0.1$

Four travelers who have made hotel reservations in this study.

This means that $n = 4$

a) What is the probability that at least two of the four selected will turn to be no-shows?

This is $P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)$

In which

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

$P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486$

$P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036$

$P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001$

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0486 + 0.0036 + 0.0001 = 0.0523$

0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) What is the most likely value for X?

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

$P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.6561$

$P(X = 1) = C_{4,1}.(0.1)^{1}.(0.9)^{3} = 0.2916$

$P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486$

$P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036$

$P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001$

X = 0 has the highest probability, which means that 0 is the most likely value for X.

7 0

3 years ago

Suppose triangle ABC has vertices at A(1, 0), B(10, 0), and C(2, 6). After a 60° counterclockwise rotation about the origin, ver

Artemon [7]

Check the picture attached.

Let OB be the radius of circle with center O.

Let B' be the image of B after the described rotation

OB and OB' are sides of the equilateral triangle OBB'.

The x coordinate of B' is the midpoint of OB, that is 5.

In the right triangle B', point (5, 0) and B:

Distance point (5, 0) to B is 5
|B'B|=|OB|=10

so by the pythagorean theorem:

$a= \sqrt{ 10^{2} - 5^{2} } = \sqrt{ 2^{2} *5^{2} - 5^{2} }= \sqrt{5^{2}(4-1)}=5 \sqrt{3}$ units

Answer: $5 \sqrt{3}$