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2 years ago

Find an equation of the line through (4,5) and parallel to y=3x+5.

Mathematics

1 answer:

melisa1 [442]2 years ago

3 0

GIVEN: y = 3x + 5

--> slope of given line = 3

--> This means that your equation must have the same slope, or "m"

FORMULA OF A LINE: y = mx + b

--> Use point-slope form to find "b"

POINT SLOPE FORM: y - y₁ = m ( x - x₁ )

--> Plug in your given point AND the slope of the given line

y - 5 = 3 ( x - 4 )

y - 5 = 3x - 12

y = 3x - 7

<h3>ANSWER: y = 3x - 7</h3>

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A test has 20 true/false questions. What is the probability that a student passes the test if they guess the answers? Passing me

Minchanka [31]

Using the binomial distribution, it is found that:

The probability that the student will get 15 correct questions in this test by guessing is 0.0207 = 2.07%.

For each question, there are only two possible outcomes, either the guess is correct, or it is not. The guess on a question is independent of any other question, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

There are 20 questions, hence $n = 20$ .

Each question has 2 options, one of which is correct, hence $p = \frac{1}{2} = 0.5$

The probability is:

$P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

In which:

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

$P(X = 15) = C_{20,15}.(0.5)^{15}.(0.5)^{5} = 0.0148$

$P(X = 16) = C_{20,16}.(0.5)^{16}.(0.5)^{4} = 0.0046$

$P(X = 17) = C_{20,17}.(0.5)^{17}.(0.5)^{3} = 0.0011$

$P(X = 18) = C_{20,18}.(0.5)^{18}.(0.5)^{2} = 0.0002$

$P(X = 16) = C_{20,19}.(0.5)^{19}.(0.5)^{1} = 0$

$P(X = 17) = C_{20,20}.(0.5)^{20}.(0.5)^{0} = 0$

Then:

$P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0148 + 0.0046 + 0.0011 + 0.0002 + 0 + 0 = 0.0207$

The probability that the student will get 15 correct questions in this test by guessing is 0.0207 = 2.07%.

You can learn more about the binomial distribution at brainly.com/question/24863377

6 0

3 years ago

Evaluate this expression mn^2 (btw m=2 and n=3)

Molodets [167]

Answer:

529

Step-by-step explanation:

4 0

3 years ago

Ana purchased a used car with 35,500 miles on it. After 4 years of driving, the car has 90,300 miles on it

Aleksandr-060686 [28]

Your answer is this
13,700
Because
x=y

5 0

3 years ago

Two more than the Quotient of twelve and a number

chubhunter [2.5K]

Answer:

x/12+2

Step-by-step explanation:

the answer to the question

4 0

3 years ago

The managers of 21 supermarkets counted the number of cars in their parking lots on the same day. The results are shown in the l

mixer [17]

The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: $(\frac{n+1}{2}) th\ term$

Putting n=21

$(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term$

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

For First Half:

98, 100, 101, 102, 108, 109, 111,118, 129, 132

$Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5$

For Second Half:

135, 135, 145, 146, 146, 156, 170 176, 180, 180

$Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151$

Now

Interquartile Range:

$IQR = Q_3-Q_1\\= 151-108.5\\=42.5$

Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

#LearnwithBrianly

7 0

3 years ago