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erastova [34]
2 years ago
8

How would you factorise 3y + 12

Mathematics
1 answer:
Flauer [41]2 years ago
7 0

Answer:

lol it so easy man 3(y+4) is the answer

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2500

Step-by-step explanation:

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How are real numbers used to describe a real-world situations
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3 0
3 years ago
For the month of March in a certain city, 57% of the days are cloudy. Also in the month of March in the same city, 55% of the da
oksian1 [2.3K]

Answer: 0.9649

Step-by-step explanation:

Let A denote the event that the days are cloudy and B denotes the event that the days are rainy.

Given : For the month of March in a certain city, the probability that days are cloudy :P(A)=0.57

Also in the month of March in the same city,, the probability that the days are cloudy and rainy :P(A\cap B)=0.55

Now by using the conditional probability, the probability that a randomly selected day in March will be rainy if it is cloudy will be :-

P(B|A)=\dfrac{P(A\cap B)}{P(A)}

\Rightarrow\ P(B|A)=\dfrac{0.55}{0.57}\\\\=0.964912280702\approx0.9649\ \ \text{[Rounded to four decimal places.]}

Hence, the probability that a randomly selected day in March will be rainy if it is cloudy = 0.9649

4 0
3 years ago
Joshua finds the perimeter of the following composite figure composed of a square and right triangle so that a braid may be cut
Bumek [7]

The perimeter of a shape is the sum of its side lengths.

<em>Joshua miscalculated the third length of the triangle</em>

First, we calculate the third length (x) of the triangle.

Because the triangle is right-angled, we can make use of Pythagoras.

So we have:

x^2 = 7^2 + 7^2

x^2 = 2(7^2)

Take square roots of both sides

x = \sqrt{2(7^2)

x = 7\sqrt{2

The other shape in the figure is a square.

So, the perimeter (P) is

P = 7 + x + x + x + 7 ---- i.e. the sum of the visible lengths

So, we have:

P = 7 + 7\sqrt2 + 7\sqrt2 + 7\sqrt2+ 7

Evaluate like terms:

P = 14 + 21\sqrt2

Hence, Joshua's error is that:

<em>He miscalculated the third length of the triangle</em>

Read more about perimeters at:

brainly.com/question/6465134

4 0
2 years ago
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