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2 December, 11:13

the percent p of the u. s. population with internet access is given by 25p-34t=1378 where t is the number of years after 1990. what percent of the u. s. population had interent access in 2005? (round to one decimal place) in what year will 89% have internent access?

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  1. 2 December, 12:05
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    Hence, 75.52 percent of the u. s. population had internet access in 2005. After 25 years or in the year 2015 89% have internet access.

    Step-by-step explanation:

    the percent p of the u. s. population with internet access is given by:

    25 p-34 t=1378

    where t is the number of years after 1990.

    We are asked to find what percent of the u. s. population had internet access in 2005 i. e. we need to find the value of p after 15 years i. e. t=15.

    ⇒ 25p-34 * (15) = 1378

    ⇒ 25p-510=1378

    ⇒ 25p=1378+510

    ⇒ 25p=1888

    ⇒ p=75.52

    Hence, 75.52 percent of the u. s. population had internet access in 2005.

    Next we are asked to find in what year will 89% have internet access?

    i. e. we have to find the value of t when we are given p=89

    ⇒ 25*89-34t=1378

    ⇒ 2225-34t=1378

    ⇒ 2225-1378=34t

    ⇒ 847=34t

    ⇒ or 34t=847

    ⇒ t=24.9112≈25

    Hence after 25 years 89% will have internet access.

    i. e. in the year 2015 89% will have internet access.
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