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REY [17]
2 years ago
14

A researcher is interested in testing to determine if the mean price of a casual lunch is different in the city than it is in th

e suburbs. The null hypothesis is that there is no difference in the population means (i.e. the difference is zero). The alternative hypothesis is that there is a difference (i.e. the difference is not equal to zero). He randomly selects a sample of 9 lunch tickets from the city population resulting in a mean of $14.30 and a standard deviation of $3.40. He randomly selects a sample of 14 lunch tickets from the suburban population resulting in a mean of $11.80 and a standard deviation $2.90. He is using an alpha value of .10 to conduct this test. Assuming that the populations are normally distributed and that the population variances are approximately equal, the degrees of freedom for this problem are _______.
Mathematics
1 answer:
Softa [21]2 years ago
5 0

Answer:

The degrees of freedom for this problem are 21

Step-by-step explanation:

Degrees of freedom:

When testing an hypothesis involving two samples, the number of degrees of freedom is given by:

df = n_1 + n_2 - 2

In which n_1 is the size of the first sample and n_2 is the sample of the second sample.

In this question:

Samples of 9 and 14, so n_1 = 9, n_2 = 14

The degrees of freedom for this problem are

df = n_1 + n_2 - 2

df = 9 + 14 - 2 = 21

The degrees of freedom for this problem are 21

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AveGali [126]
I’m not 100% sure I’m correct on this but when she started at 12,584 miles, her tank was full. After driving, it was at 12,584 miles which is a difference of 208 miles. So if you put that into a fraction: it would be 31.12 dollars/208 miles, Which means every 31.12 dollars spent, she is able to drive 208 miles. Now all you need to do is simplify that. If you divide 31.12 by 208, you get around .149. That rounded would be about .15 gallons of gas. So every (and I don’t know if t tells you to round or not so be careful on that) .149 dollars, (not 149, it’s .149) she goes one mile.
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3 years ago
PLs help 50 PTS!!!!! PLEASE ILL GIVE BRAINLIEST!!!!!
Nookie1986 [14]

Answer:

\large\boxed{y=\dfrac{1}{4}x^2-x-4}

Step-by-step explanation:

The equation of a parabola in vertex form:

y=a(x-h)^2+k

<em>(h, k)</em><em> - vertex</em>

The focus is

\left(h,\ k+\dfrac{1}{4a}\right)

We have the vertex (2, -5) and the focus (2, -4).

Calculate the value of <em>a</em> using k+\dfrac{1}{4a}

<em>k = -5</em>

-5+\dfrac{1}{4a}=-4        <em>add 5 to both sides</em>

\dfrac{1}{4a}=1           <em>multiply both sides by 4</em>

4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4

\dfrac{1}{a}=4\to a=\dfrac{1}{4}

Substitute

a=\dfrac{1}{4},\ h=2,\ k=-5

to the vertex form of an equation of a parabola:

y=\dfrac{1}{4}(x-2)^2-5

The standard form:

y=ax^2+bx+c

Convert using

(a-b)^2=a^2-2ab+b^2

y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5

<em>use the distributive property: a(b+c)=ab+ac</em>

y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4

3 0
3 years ago
23n=2.4<br> A. 3.6 <br><br> B. 7.2 <br><br> C. 1.2<br><br> D. 1.6
svp [43]

\huge\text{Hey there!}

\mathsf{23n =2.4}

\large\text{DIVIDE 23 to BOTH SIDES}

\mathsf{\dfrac{23n}{23}=\dfrac{2.4}{23}}

\large\text{Cancel out: }\mathsf{\dfrac{23}{23}}\large\text{ because it gives you 1}

\large\text{Keep: }\mathsf{\dfrac{2.4}{23}}\large\text{ because it helps us solve for n}

\mathsf{n = \dfrac{2.4}{23}}

\mathsd{\dfrac{2.4}{23}= 2.4\div 23\rightarrow \bf 0.104348}

\boxed{\boxed{\large\textsf{Answer: \huge n = \bf 0.104348}}}\huge\checkmark

\boxed{\textsf{NONE OF THE ABOVE is your ANSWER}}

\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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lubasha [3.4K]

Answer:

7a

Step-by-step explanation:

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3 years ago
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Removing which point from the coordinate plane would make the graph a function of X?
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You would need to remove (-2, 1)
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