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Finger [1]
3 years ago
10

PLEASE HELP MEE ITS EXTREMELY URGENT

Mathematics
1 answer:
Savatey [412]3 years ago
5 0

Answer:

x = pm - pn

Step-by-step explanation:

m = n +  \frac{x}{p}  \\  \\ m - n =  \frac{x}{p}  \\  \\ p(m - n) = x \\  \\  \huge \red { \boxed{x = pm - pn}}

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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
4 years ago
2m·4b−3a·2n−0.2b·5m+n·5a−5bm+8an
cupoosta [38]

Answer:

2bm + 7an

Step-by-step explanation:

Multiply.

8bm - 6an - bm + 5an - 5bm + 8an

Combined like terms:

8bm - <em>6an</em> - bm + <em>5an</em> - 5bm + <em>8an</em>

= 2bm + 7an

8 0
3 years ago
Can you help me with this??
siniylev [52]
8*8 - 4*4*pi*(1/2) = 64 - 8*pi
6 0
4 years ago
What are the values of u and v
lubasha [3.4K]

Answer:

u=72

v=61

Step-by-step explanation:

for U:

54+54=108

180-108=72

For V:

180-58=122

122/2=61

5 0
3 years ago
Lisa spends part of her year as a member of a gym. She then finds a better deal at another gym
Tema [17]
I’m having trouble understanding this question.?
5 0
3 years ago
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