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hodyreva [135]
2 years ago
13

Evaluate -29 - X when x = 6

Mathematics
1 answer:
notka56 [123]2 years ago
6 0

Answer:

wouldn't that just be -35?

Step-by-step explanation:

you just have to do -29-6

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The amount a spring will stretch varies directly with the amount of weight attached to the spring. If a spring stretches 2.1 inc
aleksklad [387]

Answer:

1.05\ in

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form y/x=k or y=kx

Let

x -----> the amount of weight

y ----- the amount a spring will stretch

we know that

For x=80 lb, y=2.1 in

<em>Find the value of k (constant of proportionality)</em>

k=y/x

substitute the values

k=2.1/80

The linear equation is

y=(2.1/80)x

How far will it stretch when 40 pounds is attached?

For x=40 lb

substitute the value of x in the linear equation and solve for y

y=(2.1/80)(40)=1.05\ in

3 0
3 years ago
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iVinArrow [24]

This is a right triangle.  If the triangle were laid down, on an end, a right angle would be visible.

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2 years ago
Albert multiplied a whole number by a fraction. The whole number is greater than one. The fraction is greater than zero and less
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What are the answer options?

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3 years ago
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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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