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Ilya [14]
3 years ago
14

A quantity p varies jointly with r and s. Which expression represents the constant of variation, k?

Mathematics
2 answers:
natima [27]3 years ago
6 0

Answer:

k=\frac{p}{rs}

Second option is correct.

Step-by-step explanation:

We have been given that p varies jointly with r and s.

Thus, we can write it mathematically as

p=krs

Here, k is the constant of variation.

Now, we have to find the value of k.

Divide the above equation by rs

\frac{p}{rs}=\frac{krs}{rs}\\\\k=\frac{p}{rs}

Hence, the value of k is k=\frac{p}{rs}

Second option is correct.

Zina [86]3 years ago
3 0

The correct answer is p/rs.

In order to find this first state the direct variation with the constant k included.

p = krs

Now to find the value of k, we simply divide away the r and s

p/rs = k

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Step-by-step explanation:

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7 0
3 years ago
If the ratio of a circle's sector to its total area is 3/8, what is the measure of its sector's arc?
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Imagine the area is 360 degrees, or 2\pi radians.

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Which number is an irrational number?
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For this case we have by definition, that an irrational number is a number that can not be expressed as a fraction \ frac {a} {b}, where a and b are integers and b is different from zero.

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Accuracy in taking orders at a drive-through window is important for fast-food chains. Periodically, QSR Magazine publishes "The
pav-90 [236]

Answer:

a) 0.7412 = 74.12% probability that all the three orders will be filled correctly.

b) 0.0009 = 0.09% probability that none of the three will be filled correctly

c) 0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d) 0.9991 = 99.91% probability that at least one of the three will be filled correctly

e) 0.0082 = 0.82% probability that only your order will be filled correctly

Step-by-step explanation:

For each order, there are only two possible outcomes. Either it is filled correctly, or it is not. Orders are independent. This means that we use the binomial probability distribution to solve this question.

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.

The percentage of orders filled correctly at Burger King was approximately 90.5%.

This means that p = 0.905

You and 2 friends:

So 3 people in total, which means that n = 3

a. What is the probability that all the three orders will be filled correctly?

This is P(X = 3).

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

P(X = 3) = C_{3,3}.(0.905)^{3}.(0.095)^{0} = 0.7412

0.7412 = 74.12% probability that all the three orders will be filled correctly.

b. What is the probability that none of the three will be filled correctly?

This is P(X = 0).

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

P(X = 0) = C_{3,0}.(0.905)^{0}.(0.095)^{3} = 0.0009

0.0009 = 0.09% probability that none of the three will be filled correctly.

c. What is the probability that one of the three will be filled correctly?

This is P(X = 1).

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

P(X = 1) = C_{3,1}.(0.905)^{1}.(0.095)^{2} = 0.0245

0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d. What is the probability that at least one of the three will be filled correctly?

This is

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

With what we found in b:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0009 = 0.9991

0.9991 = 99.91% probability that at least one of the three will be filled correctly.

e. What is the probability that only your order will be filled correctly?

Yours correctly with 90.5% probability, the other 2 wrong, each with 9.5% probability. So

p = 0.905*0.095*0.095 = 0.0082

0.0082 = 0.82% probability that only your order will be filled correctly

7 0
3 years ago
Can someone please teach me how to do this? You can please do like 2 questions and please explain exactly how you got the answer
spayn [35]

All of these questions require one thing: trigonometric functions.

There are 3 main trigonometric functions, which can only be used on right triangles: sine, cosine, and tangent.

Sine = opposite / hypotenuse

Cosine = adjacent / hypotenuse

Tangent = opposite / adjacent

When trying to figure out what function to use, we always start by looking from the angle. Take problem a, for example. Looking from angle E, of which the value is not given, we have the side opposite and the side adjacent. Therefore, we should use the tangent function.

---The hypotenuse is always the longest side of the triangle. It is never considered the opposite or adjacent side.

Let's set up our function with the given information from problem a.

tan(x) = 9.7 / 5.2

---The tangent of an unknown angle is equal to the quotient of the opposite side and the adjacent side.

Now, solving for the value of x will require a calculator. We'll need to use what's called an inverse trigonometric function. Most calculators have these directly above the regular trigonometric functions, and the inverse functions are accessed using a "second" key.

---Ensure that your calculator is in degrees, not radians!

x = tan^-1(9.7 / 5.2)

x = 61.805 = 62 degrees

Next, let's take a look at problem b. This time, we're solving for a side length instead of an angle. But, we're still going to start by looking from our angle.


Looking from the 38 degree angle, we are given the adjacent side and an unknown hypotenuse. Therefore, we should use the cosine function.

cosine(38) = 53.1 / r

---The cosine of a 38 degree angle is equal to the quotient of 53.1 and an unknown hypotenuse, r.

Use your algebra skills to isolate the variable r.

r * cosine(38) = 53.1

r = 53.1 / cosine(38)

---From here, all you need to do is plug this into your calculator. Since we are solving for a side length (and given an angle), we are just using the regular trigonometric function buttons on the calculator.

r = 67.385 = 67.4 units

Hope this helps!

4 0
2 years ago
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