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3 years ago

49.8751 correct to 4 significant figures

Mathematics

2 answers:

dlinn [17]3 years ago

8 0

Answer:

49.88

i think this is it

ella [17]3 years ago

7 0

Step-by-step explanation:

$49.8751 = 49.88$

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There are 5 red marbles, 8 blue marbles, and 12 green marbles in a bag.

cricket20 [7]

To find the percentage always multiply whatever decimal number you have by 100% 
1) 5 + 8 + 12 = 25 (total number in bag)
5/25 x 100% (because you're trying to find probability of red)
= 20%

2) 5 + 8 = 13 (# of red AND blue)
13/25 x 100% (divide out of the total)
=52%

3) 5/25 = 0.2 x 12/25 = 0.48 (find probability of drawing red and green)
0.2 x 0.48 x 100%
= 9.6%

4) 10 + 12 + 28 = 50 (to find total number of trials)
28/50 x 100%
= 56%

5) 5 x 7 (number of sections on first spinner multiply by # of sections on second spinner)
= 35Its B 20%

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4 years ago

Read 2 more answers

Write an equation of the line that passes through (0,5) and (4,5). plssss

Alenkasestr [34]

I believe it would be y=0x + 5

4 0

3 years ago

A baseball pitcher won 60% of his games. How many did he win if he pitched 35 games?

Sergio039 [100]

Answer:

21 games

Step-by-step explanation:

Multiply the number of games by the percentage.

35 · 0.60 = 21

6 0

3 years ago

If 223x=78,find x. If 223x=78, find x

Maslowich

Answer:0.3273542601

Step-by-step explanation:

73÷223=0.3273542601

6 0

4 years ago

I don’t know how to solve this

Lady_Fox [76]

Answer:

$\theta =2\pi k,\ \ k\in Z\ \\\text{or}\ \\\theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

Step-by-step explanation:

Given:

$\cos \theta-\sqrt{3}\sin \theta=1$

Divide this equation by 2:

$\dfrac{1}{2}\cos \theta-\dfrac{\sqrt{3}}{2}\sin \theta=\dfrac{1}{2}$

Note that

$\cos \dfrac{\pi }{3}=\dfrac{1}{2}\\ \\\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$

So, the previous equation is

$\cos \dfrac{\pi}{3}\cdot \cos \theta-\sin \dfrac{\pi}{3}\cdot \sin \theta=\dfrac{1}{2}$

Remind that

$\cos x\cos y-\sin x\sin y=\cos (x+y),$

then

$\cos \left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{2}$

The solution of this equation is

$\dfrac{\pi}{3}+\theta=\pm \arccos \dfrac{1}{2}+2\pi k,\ \ k\in Z\\ \\\dfrac{\pi}{3}+\theta=\pm \dfrac{\pi}{3}+2\pi k,\ \ k\in Z\\ \\\theta =2\pi k,\ \ k\in Z\ \text{or}\ \theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

8 0

3 years ago