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Volgvan
2 years ago
5

Simplify 3 x (7 - 2) + 6

Mathematics
1 answer:
erastovalidia [21]2 years ago
5 0

Answer: 1

Step-by-step explanation:

8/-4 ÷ -3/9

8/-4 × 9/ -3 = 6

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What is the length of line RS? Use the law of sines to find the answer.Round to the nearest tenth.
Verdich [7]
Using the law of sines,
sin80°/3.1 = sin∠QSR/2.4
∠QSR = sin^-1(2.4sin80°/3.1)
≈ 49.679°
∠SQR = 180° - 80° - 49.679° (angles in a triangle)
= 50.321°
sin50.321°/RS = sin80°/3.1
RS = (3.1sin50.321°/sin80°)
= 2.4 units (nearest tenth)
The length of line RS is 2.4 units.
8 0
3 years ago
A sector with an area of \goldE{48\pi,\text{cm}^2}48πcm 2 start color #a75a05, 48, pi, start text, c, m, end text, squared, end
erik [133]

Answer:

3/8 π radians

Step-by-step explanation:

The Area of a sector when then central angle is in radians = 1/2r² θ

Where

θ = central angle = ?

r = 16 cm

Area of the sector = 48πcm²

Hence

Central angle = Area of a sector ÷ (1/2r²)

= 48πcm² ÷ (1/2 × 16²)

= 48πcm² ÷ 128

Central angle = 3/8π radians

Therefore, Central angle = 3/8π radians

7 0
3 years ago
Solve for x, y and z. PLEASE SOLVE ASAP I WILL GIVE 50 POINTS
Stells [14]

Answer:

x = 27/5

y = 48/5

z = 36/5

7 0
3 years ago
PLEASE HELP WITH THIS PLEASE.
Kryger [21]

Answer:

C

Step-by-step explanation:

(a^2+b^2)^2=a^2+b^2

3 0
2 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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