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Phoenix [80]
3 years ago
5

A box in a supply room contains 21 compact fluorescent lightbulbs, of which 7 are rated 13-watt, 8 are rated 18-watt, and 6 are

rated 23-watt. Suppose that three of these bulbs are randomly selected. (Round your answers to three decimal places.) (a) What is the probability that exactly two of the selected bulbs are rated 23-watt
Mathematics
1 answer:
RoseWind [281]3 years ago
3 0

Answer:

The probability that exactly two of the selected bulbs are rated 23-watt is:

= 19%

Step-by-step explanation:

Number of fluorescent light bulbs in the box = 21

7/21 = 13-watt rated light bulbs

8/21 = 18-watt rated light bulbs

6/21 = 23-watt rated light bulbs

The probability of selected a 23-watt rated light bulb = 6/21 = 28.57%

Size of random selection of bulbs = 3

The probability that 2 out of these 3 bulbs are 23-watt rated light bulbs is:

= 6/21 * 2/3 = 19%

b) The probability of selecting 2 of the 3 bulbs from exactly the 23-watt rated bulbs is the probability of the 23-watt rated bulbs being selected given the chance of selecting 2/3.

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3 3/4 ÷ (1 2/3 + 2 1/2)​
ZanzabumX [31]

Answer:

9/10.

Step-by-step explanation:

3 3/4 ÷ (1 2/3 + 2 1/2)​

= 3 3/4 / ( 5/3 + 5/2)

= 3 3/4 / ( 10/6 + 15/6)

= 3 3/4 / 25/6

= 15/ 4 / 25/6

= 15/4 * 6/25

= 90/100

= 9/10.

4 0
2 years ago
Two number cubes are rolled. Each number cube has sides numbered 1 through 6.
olasank [31]
P(odd) or P(multiple of 5)
When we roll 2 number cubes

all possible outcomes of their sum are 2,3,4,5,6,7,8,9,10,11,12.(11 possible outcomes)

Out of those possible outcomes
3,5,7,9,11 are odd (5 outcomes) and ...

5 and 10 are multiples of 5 (2 outcomes)

Now, P(odd) or P(multiple of 5) really means P(odd) + P (multiple of 5) =
(If we had “and” instead of “or” we multiply)
= (5/11) +(2/11)
=7/11
7 0
3 years ago
Read 2 more answers
What is the length of missing side JK? Round Answer to nearest tenth please.
zvonat [6]

Given:

Triangle LJK.

LJ = 89 in, LK = 28 in and m∠L = 42°

To find:

The length of missing side JK.

Solution:

LJ = k = 89

LK = j = 28

JK = l = ?

Using law of cosine:

l^{2}=j^{2}+k^{2}-2 jk \cdot \cos L

Substitute the given values.

l^{2}=28^{2}+89^{2}-2 \cdot 28 \cdot 89 \cdot \cos 42^\circ

l^{2}=784+7921-4984 \cdot (0.7)

l^{2}=784+7921-4984 \cdot (0.7431)

l^{2}=784+7921-3703.6

l^{2}=5001.4

Taking square root on both sides.

l=70.7

The length of the missing side is 70.7 in.

4 0
3 years ago
Solve the quadratic by factoring 4x2+x=-7x-3
DedPeter [7]

Answer:

x = -3/2

x = -1/2

Step-by-step explanation:

4x^2 + 8x + 3

4x^2 + 2x + 6x + 3

2x(2x+1) + 3(2x+1)

(2x + 3) (2x + 1)

x = -3/2

x = -1/2

4 0
3 years ago
Please help! How on earth do I do this? In the year 2015, Anna bought a new car for $36,000. In 2017, she was told that the valu
Klio2033 [76]

Part A

Purchase Value of Anna's Car in 2015 = $36,000

Depreciated Value of Anna's Car in 2017 = $25,000

We know that the value of the car is depreciating linearly.


From the above data, we can see that the value of the car has reduced by $36,000-$25,000 = $11,000 over 2 years.

⇒ Per year the value has been depreciated by \frac{11000}{2}

⇒ Per year the value has been depreciated by $5,500

So if the value of a car in 2015 is V₀ then value of the car after t years can be determined by the below function:

V (t) = V₀ - (t*5,500), where t indicates the years passed since 2015

OR V (t) = 36,000 - (t*5,500), where t indicates the years passed since 2015

Part B

Suggested value of Anna's car for trade in option in 2018 = $15,000

In 2018, 3 years would have been passed since 2015. Using the value of t = 3 in the function determined in Part A:

V(3) = V₀ - (3*5,500)

⇒ V(3) = 36,000 - (16,500)

⇒ V(3) = $19,000

From the above calculations, we can see that the value of the car in 2018 should be $19,000, however the suggested value for trade in is indicated as $15,000 which is lower than what the value of the car should be. Hence, basis linear depreciation, the $15,000 value is not fair for the car in 2018.

3 0
3 years ago
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