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Sliva [168]
3 years ago
10

18, 24, 26, 30 Mean: Median: Range:

Mathematics
1 answer:
Paha777 [63]3 years ago
7 0

Answer:

mean: 24.5

median:25

range:12

Step-by-step explanation:

i asked alexa...

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-fx-g=h solve for x​
katen-ka-za [31]

Answer:

x = -(h+g) / f

Step-by-step explanation:

-fx - g = h

-fx = h + g

x = (h + g) / -f

x = -(h+g)/f

7 0
3 years ago
Which system of linear inequalities is represented by the graph?
MrRa [10]

Im not good in math and this is my first question, my best guess is D

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5 0
3 years ago
Read 2 more answers
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
Solve for x<br> Enter the solutions from least to greatest.<br> -4x^2– 7 = -11
Pavlova-9 [17]

Answer:

x = 1, x = -1

Step-by-step explanation:

-4x² = -4

x² = 1

x = \sqrt{1}

x = ±1

3 0
3 years ago
Solve 34=j−12<br> .<br> j=<br> Find the value of j
Semenov [28]

Answer:

j=46

Step-by-step explanation:

34=j-12

add 12 to 34 to isolate j

the answer to that is 46

j=46

6 0
3 years ago
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