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daser333 [38]
3 years ago
9

Data for the estimated number of HIV cases worldwide between 1988 and 1996 suggests exponential growth in the number of cases ov

er this time period. An appropriate transformation of the data (base-ten logarithms were used) produced a strongly linear scatterplot, and a regression analysis was performed on the transformed data.
Here is the computer output:
Predictor Constant Coef SE Coef T F
-151.281 2.038 -74.22 0.000
Year 0.076577 0.001023 74.84 0.000
S= 0.00792598 R-Sq= 99.9% R-Sg (adj ) =99.9%
Note that HIV cases are in millions.
Which of the following value is closest to the value that this model predicts for the number of HIV cases in 1998?
(a) 1.72 million cases
(b) 2.18 million cases
(c) 17.2 million cases
(d) 52.5 million cases
(e) The predicted value is negative because we are extrapolating beyond the range o f the explanatory variable
Mathematics
1 answer:
Eduardwww [97]3 years ago
3 0

Answer:

d. 52.5 million cases

Step-by-step explanation:

from the data output in the question, we have the regression equation to be

log y = -151.281 + 0.076577x

y is the number of hiv cases which is in millions

x is the year of the hiv case

x = 1998

we put the value of x in the regression equation

log y = -151.281 + 0.076577(1998)

log y = -151.281 + 153.000846

log y  = 1.719846

since the result is in log form,

y=10^{1.791846}

y = 52.5 million cases

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Note: you did not provide the answer options, so I am, in general, solving this query to solve your concept, which anyways would clear your concept.

Answer:

Please check the explanation.

Step-by-step explanation:

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For example, putting x=3

3\left(3\right)-4y>5

9-4y>5

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\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}

\left(-4y\right)\left(-1\right)

4y

\mathrm{Divide\:both\:sides\:by\:}4

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Let us take the random y value that is less than 1.

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