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2 years ago

What is the equation of the circle in general form

Mathematics

2 answers:

Andre45 [30]2 years ago

5 0

Answer:

A) x^(2)+y^(2)-6x+10y+25=0

Step-by-step explanation:

i used a online calculator

adelina 88 [10]2 years ago

3 0

Answer:

x² + y² + 2x + 8y − 47 = 0 B. x²+y² − 2x − 8y

Step-by-step explanation:

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Which is bigger 3/12 or 1/4?

Aliun [14]

Answer:

they are equal

Step-by-step explanation:

4 0

2 years ago

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. A fence encloses a triangular field. There is

poizon [28]

Answer:

16 posts

Step-by-step explanation:

Sides are:

20 ft, 25 ft and 35 ft

There are following number of posts on each of the sides:

20 ft side ⇒ 20/5 = 4
25 ft side ⇒ 25/5 = 5
35 ft side ⇒ 35/5 = 7

Total number:

4 + 5 + 7 = 16 posts

6 0

2 years ago

Please help me 50÷(-5)=

Paraphin [41]

Answer:

-10

Step-by-step explanation:

8 0

1 year ago

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A state end-of-grade exam in American History is a multiple-choice test that has 50 questions with 4 answer choices for each que

Assoli18 [71]

Answer:

Q1) The student has a 0.01% probability of passing the test.

Q2) She has a 99.91% probability of passing in the test.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he gets it correct, or he gets it wrong. So we solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

For this problem, we have that:

Question 1.

There are 50 questions, so $n = 50$ .

The student is going to guess each question, so he has a $\pi = \frac{1}{4} = 0.25$ probability of getting it right.

He needs to get at least 25 question right.

So we need to find $P(X \geq 25)$ .

Using a binomial probability calculator, with $n = 50$ and $\pi = 0.25$ we get that $P(X \geq 25) = 0.0001$ .

This means that the student has a 0.01% probability of passing the test.

Question 2.

Now, we need to find $P(X \geq 25)$ with $\pi = 0.70$ . So $P(X \geq 25) = 0.9991$

She has a 99.91% probability of passing in the test.

7 0

3 years ago

Read 2 more answers

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago