Write 0.000 376 in standard form.

The average annual premium for automobile insurance in the United States is $1411. A representative sample of annual premiums fo

Harrizon [31]

Using confidence interval concepts, it is found that:

a) The point estimate is the sample mean.

b) The confidence interval is: $\overline{x} \pm T\frac{s}{n}$

c) If $1411 is between the bounds of the interval yes, otherwise no.

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At item a, it asks to provide a point estimate for the population mean, which is the sample mean $\overline{x}$

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Item b:

In the spreadsheet, it is possible to find the <em>standard deviation for the sample</em>, thus, the t-distribution is used.
The confidence interval will be given by:

$\overline{x} \pm T\frac{s}{n}$

In which:

$\overline{x}$ is the sample mean.
T is the critical value, from the t-distribution, with a two-tailed area of $\frac{1 - 0.95}{2} = 0.025$ and n-1 degrees of freedom.
s is the standard deviation for the sample.
n is the sample size.

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Item c:

We have to check if 1411 is between $\overline{x} - T\frac{s}{n}$ and $\overline{x} +T\frac{s}{n}$ .

If it is, the interval includes the National Average, otherwise, it does not.

A similar problem is given at brainly.com/question/24232455

7 0

3 years ago

Sophia's school took a field trip a total of 21 vehicles were needed for the trip some students took the bus and some students c

MissTica

9 buses and 12 cars would be needed.

4 0

3 years ago

YALL PLSSSSSSSS HELP IM SHAKING I NEED HELP PLSS

Vilka [71]

Answer:

someone is adding dont worrie

Step-by-step explanation:

8 0

3 years ago

Consider the following quadratic function Part 3 of 6: Find the x-intercepts. Express it in ordered pairs.Part 4 of 6: Find the

maksim [4K]

Answer:

The line of symmetry is x = -3

Explanation:

Given a quadratic function, we know that the graph is a parabola. The general form of a parabola is:

$y=ax^2+bx+c$

The line of symmetry coincides with the x-axis of the vertex. To find the x-coordinate of the vertex, we can use the formula:

$x_v=-\frac{b}{2a}$

In this problem, we have:

$y=-x^2-6x-13$

Then:

a = -1

b = -6

We write now:

$x_v=-\frac{-6}{2(-1)}=-\frac{-6}{-2}=-\frac{6}{2}=-3$

Part 3:

For this part, we need to find the x-intercepts. This is, when y = 0:

$-x^2-6x-13=0$

To solve this, we can use the quadratic formula:

$x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot(-1)\cdot(-13)}}{2(-1)}$

And solve:

$x_{1,2}=\frac{6\pm\sqrt{36-52}}{-2}$ $x_{1,2}=\frac{-6\pm\sqrt{-16}}{2}$

Since there is no solution to the square root of a negative number, the function does not have any x-intercept. The correct option is ZERO x-intercepts.

Part 4:

To find the y intercept, we need to find the value of y when x = 0: