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nata0808 [166]
3 years ago
12

Find the correct value of x

Mathematics
1 answer:
bearhunter [10]3 years ago
3 0

Answer: x = 20

Step-by-step explanation:

Right angle = 90

90 - 47 = 43

43 - 3 = 40

40/2 = 20

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alexandr402 [8]
3x^4 becomes 3(3x)^4=3*3^4*x^4
y^2 becomes (3y)^2=3^2y²
the new z=3² the old z, so b is correct.
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In a video game, a guppy that escapes a net turns into three goldfish. Each goldfish can turn into two betta fish. Each betta fi
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What is 5 + 2/9 + 4 + 11/13 please and thank you I need help
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10.0683760684

Step-by-step explanation:

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What is the length of AB?<br> O A. 9<br> O B. 18<br> O C. 48<br> D. 6
insens350 [35]
<h3>Answer:  A.  9</h3>

=====================================================

Explanation:

Draw in the segments AO and OC.

Triangle ABO is congruent to triangle CBO. We can prove this through the use of the HL theorem. HL stands for hypotenuse leg.

Since the triangles are congruent, this means the corresponding pieces AB and BC are the same length.

Then we can say:

AB+BC = AC .... segment addition postulate

AB+AB = AC .... plug in BC = AB

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In short, the chord AC is bisected by the perpendicular radius drawn in the diagram. So all we do is cut AC = 18 in half to get AB = 9.

6 0
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A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday�s mail. In actuali
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Answer:

P(Y = 0) = 0.09

P(Y = 1) = 0.4

P(Y = 2) = 0.32

P(Y = 3) = 0.19

Step-by-step explanation:

Let the events be:

W = Wednesday

T = Thursday

F = Friday

S = Saturday

Their corresponding probabilities are

P(W) = 0.3\\P(T) = 0.4\\P(F) = 0.2\\P(S) = 0.1

Since Y = number of days beyond Wednesday that it takes for both magazines to arrive(so possible Y values are 0, 1, 2 or 3)

The possible number of outcomes are therefore 4^2 = 16\\(W, W), (W, T), (W, F), (W, S)\\(T, W), (T, T), (T, F), (T, S)\\(F, W), (F, T), (F, F), (F, S)\\(S, W), (S, T), (S, F), (S, S)

The values associated for each of the outcomes are as follows:

Y(W, W) = 0, Y(W, T) = 1, Y(W, F) = 2, Y(W, S) = 3\\Y(T, W) = 1, Y(T, T) = 1, Y(T, F) = 2, Y(T, S) = 3\\Y(F, W) = 2, Y(F, T) = 2, Y(F, F) = 2, Y(F, S) = 3\\Y(S, W) = 3, Y(S, T) = 3, Y(S, F) = 3, Y(S, S) = 3

The probability mass function of Y is,

P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19

7 0
3 years ago
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