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jenyasd209 [6]
3 years ago
14

A product comes in cans labeled "38 oz". A random sample of 10 cans had the following weights:

Mathematics
1 answer:
KiRa [710]3 years ago
7 0

Answer:

b. 35.9, 39.1 ounces

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

barX=37.465 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=2.27 represent the sample standard deviation

n=10 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

barX +- t*[(s)/(sqrt{n))]   (1)

In order to calculate the critical value t we need to find first the degrees of freedom, given by:

df=n-1=10-1=9

Since the Confidence is 0.95 or 95%, the value of alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that t_(alpha/2)=2.26

Now we have everything in order to replace into formula (1):

37.465-2.26[(2.27)/(sqrt{10))]=35.9    

37.465+2.26[(2.27)/(sqrt{10))]=39.1

So on this case the 95% confidence interval would be given by (35.9;39.1)    

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"point"

Step-by-step explanation:

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8 0
2 years ago
Please help me!<br> Find the number x such that f(x) =1
STatiana [176]

Answer:

D

Step-by-step explanation:

We have the piecewise function:

f(x) = \left\{        \begin{array}{ll}            -\frac{1}{2}x-1 & \quad x \leq -2 \\            x & \quad x > -2        \end{array}    \right.

And we want to find x such that f(x)=1.

So, let's substitute 1 for f(x):

1 = \left\{        \begin{array}{ll}            -\frac{1}{2}x-1 & \quad x \leq -2 \\            x & \quad x > -2        \end{array}    \right.

This has two equations. So, we can separate them into two separate cases. Namely:

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Let's solve for x in each case.

Case I:

We have:

1=-\frac{1}{2}x-1

Add 1 to both sides:

2=-\frac{1}{2}x

Let's cancel out the fraction by multiplying both sides by -2. So:

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The right side cancels:

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Flip:

x=-4

So, x is -4.

Case II:

We have:

1=x

Flip:

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This is the solution for our second case.

So, we have:

x_1=-4\text{ or } x_2=1

Now, can check to see if we have to to remove solution(s) that don't work.

Note that x=-4 is the solution to our first equation.

The first equation is defined only if x is less than -2.

-4 <em>is </em>less than -2. So, x=-4 is indeed a solution.

x=1 is the solution to our second equation.

The second equation is defined only if x is greater than or equal to -2.

1 <em>is</em> greater than or equal to -2. So, x=1 is <em>also </em>a solution.

Therefore, our two solutions are:

x_1=-4\text{ or } x_2=1

Out of our answer choices, we can pick D.

And we're done!

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