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2 years ago

Which graph represents equation (x-4)^2+(y+2)=16 F H G I

Mathematics

1 answer:

lina2011 [118]2 years ago

5 0

<h3>Answer: Choice G (lower left corner)</h3>

Why is this? Recall that (x-h)^2+(y-k)^2 = r^2 is the general template of a circle. The center being (h,k) and the radius is r.

For the equation (x-4)^2+(y+2)^2 = 16, we have a center of (4,-2) and radius 4. Note that 4^2 = 16.

The graph of choice G shows these properties so that's why it's the answer. Keep in mind that each tickmark represents 2 units. Choice I is close since the center is correct but the radius is too small.

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You have the numbers 1-24 written on slips of paper. If you choose one slip at random, what is the probability that you will not

ra1l [238]

The probability of selecting a number not divisible by 3 is: 2/3

Step-by-step explanation:

There are two methods to solve the question.

We can find the probability of numbers not divisible by 3
We can find the probability of numbers divisible by 3 and then find the complement of it

We will use the second method:

Given:

There are 24 slips

S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}

n(S) = 24

Let A be the event that the slip number is divisible by 3

Then

A = {3,6,9,12,15,18,21,24}

n(A) = 8

The probability of number divisible by 3 is:

$P(A) = \frac{n(A)}{n(S)}\\=\frac{8}{24}\\=\frac{1}{3}$

The sum of the probability of an event's occurrence and non-occurrence is 1. So the probability of numbers divisible by 3 will be subtracted from 1 to find the probability of selecting a number not divisible by 3.

The probability of selecting a number not divisible by 3 will be:

$=1-\frac{1}{3}\\=\frac{3-1}{3}\\=\frac{2}{3}$

The probability of selecting a number not divisible by 3 is: 2/3

Keywords: Probability

Learn more about probability at:

#LearnwithBrainly

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3 years ago

How can the Pythagorean Identity be used to find sin θ, cos θ, or tan θ and the quadrant location of the angle?

Archy [21]

First lets see the pythagorean identities

$sin^2 \Theta + cos^2 \Theta =1$

So if we have to solve for sin theta , first we move cos theta to left side and then take square root to both sides, that is

$sin^2 \Theta = 1-cos^2 \Theta => sin \theta = \pm \sqrt{1-cos^2 \Theta}$

Now we need to check the sign of sin theta

First we have to remember the sign of sin, cos , tan in the quadrants. In first quadrant , all are positive. In second quadrant, only sin and cosine are positive. In third quadrant , only tan and cot are positive and in the last quadrant , only cos and sec are positive.

So if theta is in second quadrant, then we have to positive sign but if theta is in third or fourth quadrant, then we have to use negative sign .

5 0

2 years ago

Need answers pls I am struggling!

Lady bird [3.3K]

Answer:

(1)....27

(2)...30

ans...............

6 0

2 years ago

If 3/2 divided by 1/4 = n, then n is between?

Alborosie

3/2 ÷ 1/4 = 3/2 x 4 = 6

4 0

3 years ago

A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o

o-na [289]

Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

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.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that $\mu = \frac{n}{60}$ , in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

$n = 25$ , and thus, $\mu = \frac{25}{60} = 0.4167$

This probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408$

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that $n = 6$

Each of these days, 0.6592 probability of no accident on this stretch, which means that $p = 0.6592$ .

This probability is P(X = 2). So

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

$P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879$

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

$\mu = 0.6\frac{n}{60} = 0.01n$

80 miles:

This means that $n = 80$ . So

$\mu = 0.01(80) = 0.8$

The probability of no damaging accidents is P(X = 0). So

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

6 0

2 years ago