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Anton [14]
3 years ago
6

In a city, 22 coffee shops are randomly selected, and the temperature of the coffee sold at each shop is noted. Use Excel to fin

d the 90% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed. Round your answers totwo decimal places and use increasing order. Temperature 151 158 163 155 155 144 140 148 145 146 158 143 152 167 163 147 170 168 176 164
Mathematics
1 answer:
forsale [732]3 years ago
7 0

Answer:

The confidence interval is (152.03, 159.77)

Step-by-step explanation:

The formula for calculating the confidence interval for a population mean is given by:

\bar{x}\pm \frac{z_{\alpha/2}*s}{\sqrt{n}}

The sample size is actually n = 20.

The sample average is \bar{x}=155.90.

Using excel to calculate the standard deviation we get s = 10.52

The confidence level is 1-\alpha=0.90 therefore \alpha=0.10

We obtain the critical value z_{\alpha/2}=1.65

In Excel I calculated the margin error before calculating the confidence interval. The margin error is given by:

\frac{z_{\alpha/2}*s}{\sqrt{n}}=\frac{1.65*10.52}{\sqrt{20}}=3.87

Now we can calculate the confidence interval.

\bar{x}\pm \frac{z_{\alpha/2}*s}{\sqrt{n}}=155.90\pm3.87=(152.03, 159.77)

Download xlsx
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PLZ HELP!!! Use limits to evaluate the integral.
Marrrta [24]

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Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

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where 1\le i\le n. Each interval has length \Delta x_i=\frac{2-0}n=\frac2n.

At these sampling points, the function takes on values of

f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}

We approximate the integral with the Riemann sum:

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3

Recall that

\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4

so that the sum reduces to

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}

Just to check:

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Answer:

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3 years ago
For what real value of $v$ is $\frac{-21-\sqrt{301}}{10}$ a root of $5x^2+21x+v$?
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Answer:

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is the value for which

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Step-by-step explanation:

Given that

x = (-21 - √301)/10 .....................(1)

is a root of the quadratic equation

5x² + 21x + v = 0 ........................(2)

We want to find the value of v foe which the equation is true.

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The equation is then

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