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3 years ago

6

the population of a country double every 10 years from 1950 to 1980.what was the percentage increase in population during this t

ime

1 answer:

fredd [130]3 years ago

3 0

700% increase.

Step-by-step explanation:

The duration was 3 decades, or 30 years. The population starts at 100%, then doubles to 200% (decade 1), then 400% (decade 2), then 800% (decade 3).

800% - 100% = 700%

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The diagram shows a stage in the construction of an angle bisector. Which sentence tells you how to perform the next step in the

Feliz [49]

Answer:

Draw intersecting arcs inside the angle from points D and E without changing the compass width.

Step-by-step explanation:

The next step from here, will be :

Draw intersecting arcs inside the angle from points D and E without changing the compass width.

After this, draw a straight line from B to the arcs made from above step. This line is the angle bisector.

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rachel bought 1 3/5 pounds of peaches for $.60 and 3 1/2 pounds of apples for $1.75. which fruit costs more per pound ? explain

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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

I don’t understand this question.

Anastaziya [24]

<span>In a triangle the sum of two sides is always greater than the third side.

5.6-4.0 < x < 5.6+4.0
1.6 < x < 9.6 </span>← answer

4 0

3 years ago