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To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine fo

Anni [7]

Answer:

The 90% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution 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 = 9 - 1 = 8

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 1.8595

The margin of error is:

M = T*s = 1.8595*2.7 = 5

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 25.3 - 5 = 20.3 milligrams

The upper end of the interval is the sample mean added to M. So it is 25.3 + 5 = 30.3 milligrams.

The 90% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.

7 0

3 years ago

What do I for for c? PLEASE HELP

ira [324]

Just draw one empty circle for 1 and then a filled circle for -1 so then it's 1-1 which equals 0

5 0

3 years ago

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term $x^2$ , which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression: $x_v=\frac{-b}{2a}$

Since in our case $a=1$ and $b=5$ , we get that the x-position of the vertex is: $x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

$y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}$

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the $x_v=-\frac{5}{2}$ . Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

$f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0$

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

5 0

3 years ago

How do I solve 2x+3y=7, x-y=1

Fittoniya [83]

<span>2x+3y=7
x-y=1

</span>2x+3y=7
x=1+y

2*(1+y)+3y=7
x=1+y

2+2y+3y=7
x=1+y

5y=5 |(:5)
x=1+y

y=1
x=2

CHECK:
2x+3y=7
x-y=1

2*2 + 3*1 = 7

x-y = 1
2-1 = 1

5 0

4 years ago

Read 2 more answers

Give two equivalent ratios of 8 : 18

m_a_m_a [10]

Possible answer

1. 4:9

2. 16:36

8 0

4 years ago

Read 2 more answers

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