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3 years ago

Suppose y varies directly with x, and y = 12 when x = 2. What direct variation

Mathematics

1 answer:

neonofarm [45]3 years ago

4 0

Answer:

$y=24$

Step-by-step explanation:

Given:

$y=kx$

Substitute:

$\left \{ {{y=12} \atop {x=2}} \right.$ into $y=kx$

$12=k\times2$

Swap the sides:

$k\times2=12$

Divide both sides of the equation by the coefficient of variable:

$k=\frac{12}{2}$

Cross out the common factor:

$k=6$

Find the equation of the function:

$y=6x$

Substitute:

$y=6\times4$

Calculate the product or quotient:

$y=24$

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In the proportion x/5= 15/3 which number is equal to x in the proportion?

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Answer:
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Step-by-step explanation:
We are given a word phrase:
In the proportion x over 5 equals 15 over 3 , which number is equal to x in the proportion?
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3 years ago

Given the three vertices W(−1, 4), X(−3, −2), and Y(3, −4), what are the coordinates of Z that make quadrilateral WXYZ a square?

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Answer:

I don't see how the three existing points could ever become a square with the addition of a foiurth point.

Step-by-step explanation:

See the attached image.

A square would require that all angles be 90 degrees. The given points are the top three points on the graph. If we enter the two equations that intersect these points (blue and black lines), we can see that the angle on top is not 90 degrees. I can't see that this could ever be a square with a fourth point, z. I did find a value for z that make the four points a parallelogram.

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3 years ago

Miriam is picking out some movies to rent, and she is primarily interested in comedies and foreign films. She has narrowed down

Keith_Richards [23]

Answer:

There are 795 combinations.

Step-by-step explanation:

The number of ways or combinations in which we can select k element from a group of n elements is given by:

$nCk=\frac{n!}{k!(n-k)!}$

So, if Miriam want to choose 3 movies with at least two comedies, she have two options: Choose 2 comedies and 1 foreign film or choose 3 comedies.

Then, the number of combinations for every case are:

1. Choose 2 Comedies from the 10 and choose 1 foreign film from 15. This is calculated as:

$10C2*15C1=\frac{10!}{2!(10-8)!}*\frac{15}{1!(15-14)!}$

$10C2*15C1=675$

2. Choose 3 Comedies from the 10. This is calculated as:

$10C3=\frac{10!}{3!(10-3)!}=120$

Therefore, there are 795 combinations and it is calculated as:

675 + 120 = 795

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3 years ago

Find the probability that in a pig family of 4 offspring there will be at least 1 male and 1 female. Assume that the probability

Naya [18.7K]

The probability of there being zero males is given by:
$P(zero\ males)=4C0\times0.5^{0}\times0.5^{4}=\frac{1}{16}$
Similarly the probability of there being zero females is also 1/16.
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3 years ago