All categories

2 years ago

1990 population was 254 million in 2014 it was 303 what was the percentage increase in the population

Mathematics

1 answer:

Ann [662]2 years ago

7 0

Answer:

19.29%

Step-by-step explanation:

Given

Population in 1990

= 254000000

Population in 2014

= 303000000

Increase = 303000000 - 254000000

= 49000000

% increase in population =

Increase /initial population x 100%

That’s

49000000/254000000 x 100%

0.1929 x 100%

19.29%

The population increased by 19.29%

You might be interested in

Hepppppppplppppppplppppppppp

antoniya [11.8K]

Answer:

A = 3 ft, B = 7 ft, C = 4 ft, D = 5 ft, Surface Area = 96 ft^2

Step-by-step explanation:

<u>Step 1: Determine the sides</u>

A → 3 ft

B → 7 ft

C → 4 ft

D → 5 ft

<u>Step 2: Determine the surface area</u>

Area of the sides

$A= 4\ ft * 7\ ft = 28\ ft^2$

$A= 5\ ft * 7\ ft = 35\ ft^2$

$A= 3\ ft * 7\ ft = 21\ ft^2$

Area of the top and bottom triangles

$A= \frac{h\ *\ b}{2} = \frac{3\ ft\ *\ 4\ ft}{2} = \frac{12\ ft^2}{2} = 6\ ft^2 = 2(6\ ft^2) = 12\ ft^2$

Total Area

$28\ ft^2 + 35\ ft^2 + 21\ ft^2 + 12\ ft^2$

$96\ ft^2$

Answer: A = 3 ft, B = 7 ft, C = 4 ft, D = 5 ft, Surface Area = 96 ft^2

5 0

1 year ago

Read 2 more answers

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st

Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so $\mu = 0.5, \sigma = 0.05$ .