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marusya05 [52]
4 years ago
14

What's the recommend amount of UV protection that sunglasses should provide?

Mathematics
2 answers:
BabaBlast [244]4 years ago
7 0
They should screen out 99 to 100 percent of uv rays. Hope this helps!
olga55 [171]4 years ago
6 0
Sunglasses that block 100 percent<span> of the harmful UV and have good optical quality.</span>
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There are a total of 225 pupils in Secondary one. If the number of pupils who pay their school fees through GIRO scheme is 14 ti
asambeis [7]
There are several information's already given in the question. Lat us first write them down and then go for solving the problem.

Total number of pupil <span>in Secondary one = 225
Let us assume the number of pupil who do not pay the school fees through GIRO scheme = x
Then
Number </span>of pupil who pay the school fees through GIRO scheme = 14x
So
x + 14x = 225
15x = 225
x = 15
From the above deduction, it can be concluded that 15 pupil did not join the scheme.
8 0
3 years ago
What is the length of the hypotenuse?
torisob [31]

Step-by-step explanation:

25 is the answer . hope it helps

3 0
3 years ago
Day 1 2 3 4 5
grandymaker [24]

Step-by-step explanation:

I hope this helps you

Have a great day

7 0
4 years ago
Which property of real numbers is shown below?
marusya05 [52]

Answer:

the answer is associative property of addition

Step-by-step explanation:

the things is no matter how you switch out the numbers you will have the same income.

8 0
3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
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