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hammer [34]
3 years ago
15

Water whose temperature is at 100∘C is left to cool in a room where the temperature is 60∘C. After 3 minutes, the water temperat

ure is 90∘. If the water temperature T is a function of time t given by T = 60 + 40 e k t T=60+40ekt, find the time for the water temperature to reach 65∘C. Round to the nearest hundredth of a minute.
Mathematics
1 answer:
Tju [1.3M]3 years ago
6 0

Answer:

21.68 minutes ≈ 21.7 minutes

Step-by-step explanation:

Given:

T=60+40e^{kt}

Initial temperature

T = 100°C

Final temperature = 60°C

Temperature after (t = 3 minutes) = 90°C

Now,

using the given equation

T=60+40e^{kt}

at T = 90°C and  t = 3 minutes

90=60+40e^{k(3)}

30=40e^{3k}

or

e^{3k}=\frac{3}{4}

taking the natural log both sides, we get

3k = \ln(\frac{3}{4})

or

3k = -0.2876

or

k = -0.09589

Therefore,

substituting k in 1 for time at temperature, T = 65°C

65=60+40e^{( -0.09589)t}

or

5=40e^{( -0.09589)t}

or

e^{( -0.09589)t}=\frac{5}{40}

or

e^{( -0.09589)t}=0.125

taking the natural log both the sides, we get

( -0.09589)t = ln(0.125)

or

( -0.09589)t = -2.0794

or

t = 21.68 minutes ≈ 21.7 minutes

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Marat540 [252]

Answer:

95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

Step-by-step explanation:

We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.

Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;

                          P.Q. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of residents 65-years or older who had obtained a flu shot = \frac{28}{111} = 0.25

          n = sample of residents 65-years or older = 111

          p = population proportion of residents who were getting the flu shot

<em>Here for constructing 95% confidence interval we have used One-sample z test for proportions.</em>

<u>So, 95% confidence interval for the population proportion, p is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } },\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

   = [ 0.25-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } } , 0.25+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } } ]

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Therefore, 95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

7 0
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Answer:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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