Math / Algebra

QuestionThe exponential function given by $\mathrm{H}(\mathrm{t})=80,034.64(1.0488)^{\mathrm{t}}$ , where t is the number of years after 2006, can be used to project the number of centenarians in a certain country. Use this function to project the centenarian population in this country in 2011 and in 2038.
The centenarian population in 2011 is approximately $\square$ (Round to the nearest whole number.)

Studdy Solution

STEP 1

1. The function given is $\mathrm{H}(\mathrm{t}) = 80,034.64(1.0488)^{\mathrm{t}}$ .
2. $\mathrm{t}$ is the number of years after 2006.
3. We need to find the centenarian population in 2011 and 2038.

STEP 2

1. Calculate the value of $\mathrm{t}$ for the year 2011.
2. Substitute $\mathrm{t}$ into the function to find the population in 2011.
3. Calculate the value of $\mathrm{t}$ for the year 2038.
4. Substitute $\mathrm{t}$ into the function to find the population in 2038.
5. Round the results to the nearest whole number.

STEP 3

Calculate $\mathrm{t}$ for the year 2011:
$\mathrm{t} = 2011 - 2006 = 5$

STEP 4

Substitute $\mathrm{t} = 5$ into the function:
$\mathrm{H}(5) = 80,034.64 \times (1.0488)^5$

STEP 5

Calculate $(1.0488)^5$ :
$(1.0488)^5 \approx 1.2682$

STEP 6

Calculate $\mathrm{H}(5)$ :
$\mathrm{H}(5) = 80,034.64 \times 1.2682$ $\mathrm{H}(5) \approx 101,511.68$

STEP 7

Round $\mathrm{H}(5)$ to the nearest whole number:
$\mathrm{H}(5) \approx 101,512$
The centenarian population in 2011 is approximately:
$\boxed{101,512}$

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Studdy Solution

STEP 1

1. The function given is $\mathrm{H}(\mathrm{t}) = 80,034.64(1.0488)^{\mathrm{t}}$ .
2. $\mathrm{t}$ is the number of years after 2006.
3. We need to find the centenarian population in 2011 and 2038.

STEP 2

STEP 3

Calculate $\mathrm{t}$ for the year 2011:
$\mathrm{t} = 2011 - 2006 = 5$

STEP 4

Substitute $\mathrm{t} = 5$ into the function:
$\mathrm{H}(5) = 80,034.64 \times (1.0488)^5$

STEP 5

Calculate $(1.0488)^5$ :
$(1.0488)^5 \approx 1.2682$

STEP 6

Calculate $\mathrm{H}(5)$ :
$\mathrm{H}(5) = 80,034.64 \times 1.2682$ $\mathrm{H}(5) \approx 101,511.68$

STEP 7

Round $\mathrm{H}(5)$ to the nearest whole number:
$\mathrm{H}(5) \approx 101,512$
The centenarian population in 2011 is approximately:
$\boxed{101,512}$

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Studdy Solution

STEP 1

1. The function given is H(t)=80,034.64(1.0488)t \mathrm{H}(\mathrm{t}) = 80,034.64(1.0488)^{\mathrm{t}} H(t)=80,034.64(1.0488)t.2. t \mathrm{t} t is the number of years after 2006.3. We need to find the centenarian population in 2011 and 2038.

STEP 2

STEP 3

Calculate t \mathrm{t} t for the year 2011:t=2011−2006=5 \mathrm{t} = 2011 - 2006 = 5 t=2011−2006=5

STEP 4

Substitute t=5 \mathrm{t} = 5 t=5 into the function:H(5)=80,034.64×(1.0488)5 \mathrm{H}(5) = 80,034.64 \times (1.0488)^5 H(5)=80,034.64×(1.0488)5

STEP 5

Calculate (1.0488)5 (1.0488)^5 (1.0488)5:(1.0488)5≈1.2682 (1.0488)^5 \approx 1.2682 (1.0488)5≈1.2682

STEP 6

Calculate H(5) \mathrm{H}(5) H(5):H(5)=80,034.64×1.2682 \mathrm{H}(5) = 80,034.64 \times 1.2682 H(5)=80,034.64×1.2682 H(5)≈101,511.68 \mathrm{H}(5) \approx 101,511.68 H(5)≈101,511.68

STEP 7

Round H(5) \mathrm{H}(5) H(5) to the nearest whole number:H(5)≈101,512 \mathrm{H}(5) \approx 101,512 H(5)≈101,512 The centenarian population in 2011 is approximately:101,512 \boxed{101,512} 101,512​

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1. The function given is $\mathrm{H}(\mathrm{t}) = 80,034.64(1.0488)^{\mathrm{t}}$ .
2. $\mathrm{t}$ is the number of years after 2006.
3. We need to find the centenarian population in 2011 and 2038.

Calculate $\mathrm{t}$ for the year 2011:
$\mathrm{t} = 2011 - 2006 = 5$

Substitute $\mathrm{t} = 5$ into the function:
$\mathrm{H}(5) = 80,034.64 \times (1.0488)^5$

Calculate $(1.0488)^5$ :
$(1.0488)^5 \approx 1.2682$

Calculate $\mathrm{H}(5)$ :
$\mathrm{H}(5) = 80,034.64 \times 1.2682$ $\mathrm{H}(5) \approx 101,511.68$

Round $\mathrm{H}(5)$ to the nearest whole number:
$\mathrm{H}(5) \approx 101,512$
The centenarian population in 2011 is approximately:
$\boxed{101,512}$