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Maurinko [17]
3 years ago
8

What is the relationship between the 5’s in 345,502

Mathematics
1 answer:
Wittaler [7]3 years ago
6 0
The answer is to the question is C
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A gift shop wants to make gift baskets for its regular customers. There are 24 bottles
Nuetrik [128]

Answer:

Find the biggest # that can divide all 3 #'s, and that's your answer.

Step-by-step explanation:

24 is divisible by:

1, 2, 3, 4, 6, 8, 12, 24

36 is divisible by:

1, 2, 3, 4, 6,  9, 12, 18, 36

48 is divisible by:

1, 2, 3, 4, 6, 8, 12, 16, 24, 48

The biggest # that can divide all 3 #'s is 12, so the greatest amount of baskets that can be made is 12 baskets.

6 0
3 years ago
637 x 3 find the products. Show your thinking. The first row gives some ideas for showing your thinking
Colt1911 [192]

Answer:

1,911

Step-by-step example

3 x 7

3 x 30

3 x 600

then add it all up to get the 1,911

6 0
3 years ago
I really need help with this guys please help it’s due tomorrow and I’m so frikin dumb
il63 [147K]

Possible 24 impossible 3

5 0
3 years ago
What’s 8x8 <br> This is to test how the app works not serious
lisov135 [29]

Answer:

64

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
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
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