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
gregori [183]
3 years ago
12

I WILL MARK YOU BRAINLYEST IF YOU HELP ME AND YOU SHOW WORK!!!! FOR BOTH PARTS!!!

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
6 0
A. files * photo percentage = photos. 
b. 250 * 0.12 = 30
You might be interested in
PLS HELP
myrzilka [38]

Answer:

ABD and ABE

Step-by-step explanation:


3 0
3 years ago
Read 2 more answers
How many terms are in the arithmetic sequence 13, 16, 19, ……, 70, 73?
Alexxx [7]
should be (73-13)/3=20
8 0
3 years ago
Please help me with this homework
gregori [183]
Take the square root of 10,000, which is 100.
4 0
3 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
How many $3.50 burgers can you buy for a millon dollars?
den301095 [7]
You can buy 285714 burgers with $1,000,000. Lot of burgers,eh?!?!?!?
8 0
3 years ago
Other questions:
  • Write 2 expressions for the perimeter
    10·1 answer
  • 48 oz can of shortening, total price was 2.99. what was the price per unit?
    8·1 answer
  • Find the inverse of f(x)=1/(x-3)
    5·1 answer
  • Use laws of exponents to simplify inside the parentheses of the given expression ((a ^ - 2 * b ^ 2)/(a ^ 2 * b ^ - 1)) ^ - 3
    12·2 answers
  • Eleven twelves times two thirds
    12·2 answers
  • My car payment is 312.67. My water bill is 12percent of the car payment. How much is the water bill
    6·2 answers
  • Find sin(b) in the triangle
    7·2 answers
  • Question 3 (4 points)
    14·1 answer
  • A group of 13 friends were planning a trip. On the night before they left they made a lot of phone calls. Each friend talked to
    9·1 answer
  • The lengths of the sides of a triangle are in the ratios 2:6:7
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!