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alexandr402 [8]
3 years ago
10

Simplify.

Mathematics
2 answers:
Igoryamba3 years ago
8 0
26+(-3)*(-8) \\ 
 \\ (-3)*(-8)=24 \\ 
 \\ 26+24=50 \\ 
 \\ Answer:\boxed{D.50} \\
Rasek [7]3 years ago
6 0
\rm 26+(-3)\cdot(-8)=26+24=\bold{50} \\ \\ \bold{Answer:D}
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The Environmental Protection Agency (EPA) publishes fuel economy values that are known to be good estimates of the fuel economy
r-ruslan [8.4K]

Answer:

The 90% confidence interval for the mean combined fuel economy for Ford Explorers is between 22.95 and 23.63 mpg.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 1.7531

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.7531\frac{0.78}{\sqrt{16}} = 0.34

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 23.29 - 0.34 = 22.95 mpg

The upper end of the interval is the sample mean added to M. So it is 23.29 + 0.34 = 23.63 mpg

The 90% confidence interval for the mean combined fuel economy for Ford Explorers is between 22.95 and 23.63 mpg.

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kirill [66]
Explanation

We must the tangent line at x = 3 of the function:

f(x)=(\ln x)^3.

The tangent line is given by:

y=m*(x-h)+k.

Where:

• m is the slope of the tangent line of f(x) at x = h,

,

• k = f(h) is the value of the function at x = h.

In this case, we have h = 3.

1) First, we compute the derivative of f(x):

f^{\prime}(x)=\frac{d}{dx}((\ln x)^3)=3*(\ln x)^2*\frac{d}{dx}(\ln x)=3*(\ln x)^2*\frac{1}{x}=\frac{3(\ln x)^2}{x}.

2) By evaluating the result of f'(x) at x = h = 3, we get:

m=f^{\prime}(3)=\frac{3}{3}*(\ln3)^2=(\ln3)^2.

3) The value of k is:

k=f(3)=(\ln3)^3

4) Replacing the values of m, h and k in the general equation of the tangent line, we get:

y=(\ln3)^2*(x-3)+(\ln3)^3.

Plotting the function f(x) and the tangent line we verify that our result is correct:

Answer

The equation of the tangent line to f(x) and x = 3 is:

y=(\ln3)^2*(x-3)+(\ln3)^3

6 0
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