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svetoff [14.1K]
3 years ago
15

By coincidence, Leon's Obedience School trained the same number of dogs and cats last

Mathematics
1 answer:
LuckyWell [14K]3 years ago
4 0

Answer:

14

Step-by-step explanation:

In this case LCM is needed and the LCM of 2and 7 is 14. I hope this helps.

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(cos 2theta)/(1 + sin 2theta) = (1 - tan theta)/(1 + tan theta)
andre [41]

Step-by-step explanation:

Here is the solution...hope it helps:)

3 0
1 year ago
NEED HELP FINDING SLOPE OF THE LINE
mamaluj [8]

Answer:

slope = - \frac{2}{3}

Step-by-step explanation:

calculate the slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (- 4, 2) and (x₂, y₂ ) = (2, - 2) ← 2 points on the line

m = \frac{-2-2}{2-(-4)} = \frac{-4}{2+4} = \frac{-4}{6} = - \frac{2}{3}

5 0
1 year ago
A model of a house has been drawn on a coordinate grid. One corner of the house has been drawn at (1,5). The drawing will be tra
asambeis [7]
It should me b I don’t know if it is but it should be
7 0
3 years ago
Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second car
Nikolay [14]

Answer:

probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3

Step-by-step explanation:

First of all;

Let B1 be the event that the card with two red sides is selected

Let B2 be the event that the

card with two black sides is selected

Let B3 be the event that the card with one red side and one black side is

selected

Let A be the event that the upper side of the selected card (when put down on the ground)

is red.

Now, from the question;

P(B3) = ⅓

P(A|B3) = ½

P(B1) = ⅓

P(A|B1) = 1

P(B2) = ⅓

P(A|B2)) = 0

(P(B3) = ⅓

P(A|B3) = ½

Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;

P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]

Thus;

P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]

P(B3|A) = (1/6)/(⅓ + 0 + 1/6)

P(B3|A) = (1/6)/(1/2)

P(B3|A) = 1/3

5 0
3 years ago
The sum of three different numbers is 18 if every number is a prime number what are the three numbers
Aleks04 [339]
<span>it would be 5+2+11 because a lll number are prime</span>
5 0
2 years ago
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