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eimsori [14]
3 years ago
8

Help please it would help my grade alot

Mathematics
1 answer:
Alexxandr [17]3 years ago
3 0

Answer:

It cooled 58.4 degrees

Step-by-step explanation:

T(t) = 68 + 144e ^(-.052t)

We want to find the value when t= 10

T(10) = 68 + 144e ^(-.052*10)

        = 68 +85.611

 T(10)         =153.611

We also need to know

T(0)= 68 + 144e ^(-.052*0)

       =68+144

        =212

We want to know how much it cools

That would be the temperature at  time 0 minus the temperature at  time 10


T(0) - T (10)

212-153.611

58.389

To the nearest tenth

58.4 degrees

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The first option is correct. Option A is correct.

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Critical value at 80% confidence level for sample size of 90 is obtained from the z-tables.

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Standard error of the mean = σₓ = √[p(1-p)/n]

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σₓ = √[0.3333×0.6667)/90] = 0.0496891568 = 0.04969

80% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.3333 ± (1.28 × 0.04969)

CI = 0.3333 ± 0.0636021207

80% CI = (0.2696978793, 0.3969021207)

80% Confidence interval = (0.270, 0.397)

Hope this Helps!!!

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