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sladkih [1.3K]
3 years ago
11

%2Cr%3D-1%2Cs%3D0" id="TexFormula1" title="\frac{r}{2}+\frac{s}{r} -\frac{r}{4}+\frac{1}{5},r=-1,s=0" alt="\frac{r}{2}+\frac{s}{r} -\frac{r}{4}+\frac{1}{5},r=-1,s=0" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
zloy xaker [14]3 years ago
3 0

Answer:

-1/20

Step-by-step explanation:

r/2 + s/r - r/4 + 1/5 =

= -1/2 + 0/(-1) - (-1)/4 + 1/5

= -1/2 + 1/4 + 1/5

= -10/20 + 5/20 + 4/20

= -5/20 + 4/20

= -1/20

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Evaluate the expression.<br><br> (13^2 - 12^2) ÷ 5​
otez555 [7]
Answer:

5

Step-by-step explanation:

(13^2 - 12^2) ÷ 5

(169 - 144) ÷ 5

25 ÷ 5

5
5 0
2 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
HELP I WILL MARK BRAINLIEST
Ede4ka [16]

Answer:

the answer is A.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ag
Katyanochek1 [597]

Answer:

At 5% significance level, it is statistically evident that    there is nodifference in the proportion of college students who consider themselves overweight between the two poll                                  

Step-by-step explanation:

Given that a poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ago.

Let five years ago be group I X and as of now be group II Y

H_0: p_x =p_y\\H_a: p_x \neq p_y

(Two tailed test at 5% level of significance)

                Group I            Group II              combined p

n                  270                 300                         570

favor            120                  140                          260

p                   0.4444            0.4667                   0.4561

Std error   for differene = \sqrt{\frac{0.4561(1-0.4561)}{570} } \\=0.0209

p difference = -0.0223    

Z statistic = p diff/std error =       -1.066

p value =0.2864

Since p value >0.05, we accept null hypothesis.

        At 5% significance level, it is statistically evident that    there is nodifference in the proportion of college students who consider themselves overweight between the two poll                                  

7 0
3 years ago
The area of a square game board is 144 sq. In. What's the length of the sides of the board?
Aleks [24]
Sqrt of 144 = 12 so in other words your answer is 12
3 0
3 years ago
Read 2 more answers
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