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FromTheMoon [43]
2 years ago
15

Ng

Mathematics
1 answer:
Feliz [49]2 years ago
5 0

Answer:

PRT%

Step-by-step explanation:

<em>PR</em><em>T</em>

<em>= 320 \times 2.3 \times 20 \div 100 = 1472</em>

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[(12-5). 12] ÷16=?<br>can I please get some help on this? It would be greatly appreciated
lina2011 [118]

i believe it would be 5.25

12-5=7

7*12=84

84/ 16 = 5.25

i hope this helps

8 0
3 years ago
Read 2 more answers
What is the solution to this equation x+ 3 2/3= 26
jenyasd209 [6]

Answer:

Exact form : x = 67/3

Decimal form : x = 22.33333333333...

Mixed Number form : x = 22 1/3

Step-by-step explanation: Solve for x by simplifying both sides of the equation, then isolating the variable.

I hope this helps you out. :)

5 0
3 years ago
Calculate the area of this figure. Show all work and include proper units.<br> Check photo for units
Allushta [10]

Answer:

  • 92 ft²

Step-by-step explanation:

If you complete the square, you will add a right triangle with legs 10 - 6 = 4 ft.

<u>The area of the figure is:</u>

  • A = 10² - 1/2*4*4 = 100 - 8 = 92 ft²
5 0
2 years ago
In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0
3 years ago
Two similar figures have sides in the ratio of 2:3. If a side of the smaller triangle has a length of 7, what is the length of t
dusya [7]
1) 7/2 = 3.5
2) 3.5 * 3 = 10.5
answer: 10.5 
4 0
3 years ago
Read 2 more answers
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