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MAVERICK [17]
1 year ago
13

the graph shows the height of a pendant for a certain amount of time and days happy Monday everybody there may be a proportional

relationship between the number of days and the height of the plant

Mathematics
1 answer:
Alex Ar [27]1 year ago
4 0

According to the given graph, the variables does not show a proportional relationship because they points do not form an straight line. Remember that the proportional relationship is represents by a straight line whcih passes through the origin.

<h2>Therefore, the hours and the height of snow do not show a proportional relationship.</h2>
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Write the number in standard form as a decimal
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Answer:

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Step-by-step explanation:

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3 years ago
I need help asap please!!!
Mashcka [7]
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3 years ago
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mina [271]

Oh shucks, that's crazy man

6 0
3 years ago
HELPPP YOU GET 30 POINTS AND BRAINLIEST
Alex Ar [27]

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4 0
2 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
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