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Nostrana [21]
3 years ago
9

There are 8 crackers in 1 serving, 16 crackers in 2 servings, 24 crackers in 3 servings, and so on. How many crackers are in a b

ox of 12 servings?
Mathematics
2 answers:
nydimaria [60]3 years ago
7 0
Using the pattern shown, each serving contains 8 crackers. In order to find the number of crackers in 12 servings, you will multiply the number of crackers in one serving, 8
, by the number of servings in the  box. So, 8 times 12 equals 96, therefore there are 96 crackers in a box.
 
slavikrds [6]3 years ago
5 0
1 serving = 8 crackers         (1 x 8 = 8)

2 servings = 16 crackers     (2 x 8 = 16)

3 servings = 24 crackers     (3 x 8 = 24)

12 servings = ? crackers

12 x 8 = 96

The answer is 96 crackers
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kondor19780726 [428]

Answer:

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Step-by-step explanation:

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3 years ago
The following lines are ______ <br><br> 4x + 2y = 10 y = -2x + 15
zmey [24]

Answer:

Parallel

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

given

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2y = - 4x + 10 ( divide through by 2 )

y = - 2x + 5 ← in slope- intercept form

with slope m = - 2

and

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4 0
1 year 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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elena55 [62]

Answer:

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Step-by-step explanation:

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2×2=4

6 0
3 years ago
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yaroslaw [1]

Answer:

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Step-by-step explanation:

I measured it till I got the angle or you could search it up and see what the angle is.

Hope it helped! :)

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8 0
3 years ago
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