You start at (2, 3). You move right 8 units

Age (years) Population Under 15 2600 15 - 64 16000 Over 64 4000 Calculate the child dependency ratio from the chart above. Round

Brums [2.3K]

Answer:

16.25%

=0.163 (correct to 3 decimal places)

Step-by-step explanation:

The child dependency ratio of a population is defined as the number of children (Under 15 years) divided by the working-age population (15–64 years old).

$\mathrm{ Child}\;\mathrm{ dependency}\;\mathrm{ ratio}=\dfrac{{\mathrm{ Population}\,\left( \text{Under 15} \right)}}{{\mathrm{ Population}\,\left( {15-64} \right)}}\times 100$

From the given table:

Population Under 15 years = 2600

Population of the working class (between 15-64) = 16000

Therefore:

$\mathrm{ Child}\;\mathrm{ dependency}\;\mathrm{ ratio}=\dfrac{2600}{16000}\times 100\\\\=16.25\%$

=0.163 (correct to 3 decimal places)

3 0

3 years ago

Find the hourly wage if a $ 10.00 starting wage is increased by 2.5 % each year for 9 years.

djyliett [7]

10 * (1.025)^9 = 12,48

Maybe 12.49 depending on rounding.

6 0

3 years ago

The representation that is used to display the spread of the data is

Vsevolod [243]

Answer:

The spread in data is the measure of how far the numbers in a data set are away from the mean or the median. ... You can also find the spread in the data set by using the interquartile range, which is a value that is the difference between the upper quartile value and the lower quartile value.

Step-by-step explanation:

8 0

3 years ago

A substance with a half life is decaying exponentially. If there are initially 12 grams of the substance and after 70 minutes th

77julia77 [94]

Answer: 233 min

Step-by-step explanation:

This problem can be solved by the following equation:

$A=A_{o} e^{-kt}$ (1)

Where:

$A=7 g$ is the quantity left after time $t$

$A_{o}=12 g$ is the initial quantity

$t=70 min$ is the time elapsed

$k$ is the constant of decay for the material

So, firstly we need to find the value of $k$ from (1) in order to move to the next part of the problem:

$\frac{A}{A_{o}}=e^{-kt}$ (2)

Applying natural logarithm on both sides of the equation:

$ln(\frac{A}{A_{o}})=ln(e^{-kt})$ (3)

$ln(\frac{A}{A_{o}})=-kt$ (4)

$k=-\frac{ln(\frac{A}{A_{o}})}{t}$ (5)

$k=-\frac{ln(\frac{7 g}{12 g})}{70 min}$ (6)

$k=0.00769995 min^{-1}$ (7) Now that we have the value of $k$ we can solve the other part of this problem: Find the time $t$ for $A=2 g$ .

In this case we need to isolate $t$ from (1):

$t=-\frac{ln(\frac{A}{A_{o}})}{k}$ (8)

$t=-\frac{ln(\frac{2 g}{12 g})}{0.00769995 min^{-1}}$ (9)

Finally:

$t=232.697 min \approx 233 min$

5 0

3 years ago

Suppose your car gets 28 miles per gallon of gasoline, and you are driving at 55 miles per hour. Using unit analysis, find the a

pentagon [3]

<h3>Answer:</h3>

1 27/28 ≈ 1.964 gallons/hour

<h3>Step-by-step explanation:</h3>

You want gallons in the numerator of your unit rate, but that unit is in the denominator of the mileage rate. So, the computation must involve division by 28 mpg. Hours is already in the denominator of 55 mph, so the computation will involve multiplication by that rate.

... (55 mi/h)/(28 mi/gal) = (55 mi/h)·(1 gal/(28 mi)) = 55/28 gal/h

... = 1 27/28 gal/h

7 0

4 years ago