1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
fenix001 [56]
2 years ago
15

Help me with the question of O.math​

Mathematics
1 answer:
Mashutka [201]2 years ago
6 0

Answer:

24

Step-by-step explanation:

2*1+1+2*2+1+2*3+1+2*4+1=24

You might be interested in
Ayuda porfa, quien entiende???
aksik [14]

Answer:

c

Step-by-step explanation:

the right answer is c it good

6 0
2 years ago
X/5 + 1 = 10<br><br> It’s number 11
shusha [124]

Answer:

X = 45

Step-by-step explanation:

7 0
3 years ago
Suzy can bike 40 miles in 30 minutes. How far can she go in 2 hours?​
Rama09 [41]

Answer:

160 miles

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Simplyfy 5 x a x 5 x a
prohojiy [21]

Step-by-step explanation:

5×a×5×a

collect like terms

5×5×a×a

25×a^2

25a^2

8 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
Other questions:
  • Which of the following figures has a pentagon as its base? A cube B pentagonal prism C triangular prism D rectangular prism
    14·1 answer
  • What is 1.3 times 10 to the 15th power
    9·2 answers
  • Given: CB bisects
    11·1 answer
  • 2. What is the value of x in this diagram?
    13·1 answer
  • Classify the number as natural, whole, integer, rational, and/or irrational.
    12·2 answers
  • In a 45 45 right triangle what is the frmula to find either leg of the triangle
    14·1 answer
  • Twenty stamps cost 8.40. How much does it cost to buy 5 stamps
    6·2 answers
  • Find the slope of the line that passes through the points (-3, -20) and (13, 2).​
    9·2 answers
  • Which of the following is an example of imdependent events
    15·1 answer
  • Richard rolls a fair dice 72 times.
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!