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frutty [35]
3 years ago
9

The sum of the squares of two numbers is 18. the product of the two numbers is 9. find the numbers.

Mathematics
2 answers:
OLEGan [10]3 years ago
8 0

Answer:

x=3 y=3

Step-by-step explanation:

The sum of the squares of two numbers is 18, and the product of those two numbers is 9, you just need to create an equation:

So the sum of the squares is 18, the first number will be represented as X and the second as Y:

x^{2}+ y^{2} =18

And the other one is that the product of the two numbers is 9:

xy=9

We have a system of equations here, we clear X from the first one:

x=\frac{9}{y}

And instert that value of x in the first one:

x^{2}+ y^{2} =18\\(\frac{9}{y} )^{2}+ y^{2} =18\\81=y^2(18-y^2)\\y^4-18y^2+81=0\\(y^2-9)(y^2-9)=0\\Y^2-9=0\\y^2=9\\y=3

By solving this equation we get that the first number is 3.

The second number is solved by inserting the value of Y into one of the equations, in this case we will use the second:

xy=9\\x=\frac{9}{y} \\x=\frac{9}{3} \\x=3

So we get that x and y are both 3.

zhuklara [117]3 years ago
4 0
A graph shows the two numbers are either (3, 3) or (-3, -3).

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stepan [7]

Answer:

2.38 lb bag = 4.99/lb   5.08 lb bag = 5.21/lb  The 2.38 is the better buy

Step-by-step explanation:

1.88/2.38 = 4.99

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8 0
3 years ago
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An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed
masya89 [10]

Answer:

P(X\leq 1) = 0.331

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

P(X = x) = \frac{y^xe^{-y}}{x!}

Where y represents λ (average)

y = 2.3

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

Using probability notations;

P(X\leq 1) = P(X=0) + P(X =1)

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}

P(X = 0) = \frac{1 * e^{-2.3}}{1}

P(X = 0) = e^{-2.3}

P(X = 0) = 0.10025884372

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}

P(X = 1) = \frac{2.3 * e^{-2.3}}{1}

P(X = 1) =2.3 * e^{-2.3}

P(X = 1) = 2.3 * 0.10025884372

P(X = 1) = 0.23059534055

P(X\leq 1) = P(X=0) + P(X =1)

P(X\leq 1) = 0.10025884372 + 0.23059534055

P(X\leq 1) = 0.33085418427

P(X\leq 1) = 0.331

Hence, the required probability is 0.331

6 0
3 years ago
There are 4 jacks and 13 clubs in a standard, 52-card deck of playing cards. What is the probability that a card picked at rando
ale4655 [162]

Answer:

16/52, or 4/13.

Step-by-step explanation:

First, since we know that the question is asking for the probability of a club <u>or</u> a jack, we know that we have to add the two probabilities. The first probability is that of picking a club, which is 13/52. The probability of picking a jack (be sure not to overlap; don't double count the jack of clubs) is 3/52. Adding these two gives us 13/52+3/52=16/52, which simplifies to 4/13.

3 0
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How do I solve this and what’s the answer?
Varvara68 [4.7K]

Answer:

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Mike has $54 to spend on the car for a day.

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x miles = $24.80

By cross multiplication

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x = 24.80/0.20

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