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Andrej [43]
3 years ago
8

Find the slope of line passing through the two points, (7, 1) and (-2, 1).

Mathematics
1 answer:
Mamont248 [21]3 years ago
3 0

Answer: m = 0

Step-by-step explanation:

Slope (m) = y2 - y1/,x2 - x1 : meaning increase in y divided by increase in x . it can also be written as ∆y/∆x

y1= 1, y2 = 1, x1 = 7, x2 = -2

Substitute for those values in the equation above

m = 1 -1/-2 - 7

= 0/-9

Therefore,

m = 0

The slope of the line passing through those coordinates = 0

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30517578125

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Dan invests £11000 into a savings account. The bank gives 3.9% compound interest for the first 3 years and 4.5% thereafter. How
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The initial amount of the money is £11,000 and the interest is 3.9% per year for first 3 years and then 4.5% after that. If Dan invests it for 7 years, that means the interest would be 3 years of 3.9% and 4 years of 4.5%.
The calculation would be:

total money= initial amount * interestrate1 * interest 2
total money= £11000 *(100%+3.9%)^3<span>*(100%+4.5%)^4
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Divide. Write the quotient in lowest terms.<br> 5 divide 3 1/3
Step2247 [10]

Answer:

3/2 or 1 1/2

Step-by-step explanation:

5÷3 1/3

5/1 ÷ 10/3 because you have to get it into an improper fraction form in order to divide

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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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