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3 years ago

Based on the graph, the greatest percentage of increase in accidents happens in which two years? (Visualize the

Mathematics

1 answer:

Georgia [21]3 years ago

7 0

Answer:

The largest steep increase was from 1965 to 1966, while the second largest was from 1967 to 1968

Step-by-step explanation:

A) 1965 to 1966; number of accidents increased from 250 to 320.

Percentage increase = (320 - 250)/250 × 100% = 28%

B) 1967 to 1968; number of accidents increased from 300 to 370.

Percentage increase = (370 - 300)/300 × 100% = 23.33%

C) 1968 to 1969; number of accidents increased from 370 to 380.

Percentage increase = (380 - 370)/370 × 100% = 2.7%

D) 1969 to 1970; number of accidents increased from 380 to 410.

Percentage increase = (410 - 380)/380 × 100% = 7.89%

Thus;

The largest steep increase was from 1965 to 1966, while the second largest was from 1967 to 1968

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2 years ago

During the first hour of operation the farm stand owner sold 15% of the apples he had. On the second hour he sold 20% of the rem

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Answer:

3.825 kg apples.

Step-by-step explanation:

Let x kg be the amount of apples stand owner had initially.

We have been given that during the first hour of operation the farm stand owner sold 15% of the apples he had. Then the amount of sold apples will be:

$\text{The amount of sold apples during 1st hour}=\frac{15}{100}x=0.15x$

$\text{The amount of remaining apples after 1st hour}=x-0.15x$

$\text{The amount of remaining apples after 1st hour}=0.85x$

On the second hour he sold 20% of the remaining apples. So the amount of sold apples on the second hour will be 20% of 0.85x.

$\text{The amount of sold apples during 2nd hour}=0.85x*\frac{20}{100}$

$\text{The amount of sold apples during 2nd hour}=0.85x\times 0.20$

$\text{The amount of sold apples during 2nd hour}=0.17x$

During the second hour 25.5 kg of apples were sold, so we can set up an equation as:

$25.5=0.17x$

$\frac{25.5}{0.17}=\frac{0.17x}{0.17}$

$150=x$

So, stand owner had initially 150 kg of apples.

$\text{The amount of remaining apples after 2nd hour}=0.85x-0.17x$

$\text{The amount of remaining apples after 2nd hour}=0.68x$

On the third hour he sold 25% of the remaining apples. So the amount of sold apples on the 3rd hour will be 25% of 0.68x

$\text{The amount of sold apples during 3rd hour}=0.68x*\frac{25}{100}$

$\text{The amount of sold apples during 3rd hour}=0.68x*0.25$

$\text{The amount of sold apples during 3rd hour}=0.17x$

$\text{The amount of remaining apples after 3rd hour}=0.68x-0.17x$

$\text{The amount of remaining apples after 3rd hour}=0.51x$

As remaining apples were sold 5% more each following hour.

$\text{The amount of sold apples during 4th hour}=0.51x*\frac{5}{100}$

$\text{The amount of sold apples during 4th hour}=0.51x*0.05$

$\text{The amount of sold apples during 4th hour}=0.0255x$

Let us substitute x=150 into above equation we will get,

$\text{The amount of sold apples during 4th hour}=0.0255*150$

$\text{The amount of sold apples during 4th hour}=3.825$

Therefore, 3.825 kg of apples were sold during the 4th hour.

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3 years ago