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2 years ago

A line is passing through (7,-3) and it’s slope is -1/2, white the equation in point slope form

Mathematics

1 answer:

Crank2 years ago

3 0

Equation of the line in point slope form is y + 3 = -1/2(x - 7)

Step-by-step explanation:

Step 1: Given slope of the line is -1/2 and passes through the point (x1, y1) = (7, -3). Equation in point slope form is y - y1 = m(x - x1)

⇒ y - (-3) = -1/2(x - 7)

⇒ y + 3 = -1/2(x - 7)

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A model car uses the scale 1:43. The actual length of a car is 17.2 feet. What is the length of the model car?

faust18 [17]

17.2/43= 0.4

Since the scale is 1:43, you would just divide the actual length by 43.

0.4:17.2 is congruent to 1:43.

43/17.2= 2.5

1/.04= 2.5

I hope this helps!
~kaikers

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3 years ago

What is the value of x 1. 124 2. 62 3. 112 4. 68 explain why the answer is correct

Fynjy0 [20]

Answer:

<h2>3.112 because 56+56=112</h2>

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2 years ago

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Which of these shows the complete factorization of 6x^2y^2-9xy-42

sineoko [7]

Answer:

3 • (2xy - 7) • (xy + 2)

Step-by-step explanation:

(((2•3x2) • y2) - 9xy) - 42

6x2y2 - 9xy - 42 = 3 • (2x2y2 - 3xy - 14)

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2 years ago

Which number represents the value of the quotient below?

IgorC [24]

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3 years ago

In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until

Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the hypergeometric distribution is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

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

The parameters are:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

In this problem:

There are 12 bulbs, hence N = 12.

3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.

The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182$

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

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1 year ago