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prisoha [69]
3 years ago
12

Help needed now with Math question.

Mathematics
1 answer:
saveliy_v [14]3 years ago
5 0

Answer:

is the last one

Step-by-step explanation:

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Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern
Naya [18.7K]

Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases (\mu_1-\mu_2) is [-9.132 , 23.332].

Step-by-step explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population means is given by;

                            P.Q. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~ t__n__1-_n__2-2

where, \bar X_1 = sample mean sales receipts for mail-order sales = $81.70

\bar X_2 = sample mean sales receipts for internet sales = $74.60

s_1 = sample standard deviation for mail-order sales = $18.75

s_2 = sample standard deviation for internet sales = $28.25

n_1 = size of sales receipts for mail-order sales = 7

n_2 = size of sales receipts for internet sales = 11

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} } = 25.11

<em>Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>

<em />

So, 80% confidence interval for the difference between population means, (\mu_1-\mu_2) is ;

P(-1.337 < t_1_6 < 1.337) = 0.80  {As the critical value of t at 16 degree

                                         of freedom are -1.337 & 1.337 with P = 10%}  

P(-1.337 < \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < 1.337) = 0.80

P( -1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < {(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} < 1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.80

P( (\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.80

<u>80% confidence interval for</u> (\mu_1-\mu_2) =

[ (\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } , (\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ]

= [ (81.70-74.60)-1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } } , (81.70-74.60)+1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } } ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases (\mu_1-\mu_2) is [-9.132 , 23.332].

4 0
3 years ago
Imagine if you wrapped a rope tightly around the earth at the equator. how much longer would you have to make the rope if you wa
jasenka [17]
Answer: Use the formula: 2(pi)(r + 1) - 2(pi)r

In your problem, you didn't give any measurements to use, so we can't determine an exact answer. However, the given formula above would work. 

The first term calculates the extended length of rope. The second term calculates the original length of rope. All you need to do is subtract them.

Online, you can find the radius of the Earth if you need an exact value. 3,959 miles
6 0
3 years ago
How tdo cells communicate through chemical signaling.
Iteru [2.4K]

Answer: Cells typically communicate using chemical signals. These chemical signals, which are proteins or other molecules produced by a sending cell, are often secreted from the cell and released into the extracellular space. There, they can float – like messages in a bottle – over to neighboring cells.

Step-by-step explanation:

4 0
2 years ago
Write the equation of the parabola that has its x-intercepts at (1+<img src="https://tex.z-dn.net/?f=%5Csqrt%7B5%7D" id="TexForm
VladimirAG [237]

Answer:

y=2x^2-4x-8

Step-by-step explanation:

<u>Factored form of a parabola</u>

y=a(x-p)(x-q)

where:

  • p and q are the x-intercepts.
  • a is some constant.

Given x-intercepts:

  • (1+√5, 0)
  • (1-√5, 0)

Therefore:

\implies y=a(x-(1+\sqrt{5}))(x-(1-\sqrt{5}))

\implies y=a(x-1-\sqrt{5})(x-1+\sqrt{5})

To find a, substitute the given point (4, 8) into the equation and solve for a:

\implies a(4-1-\sqrt{5})(4-1+\sqrt{5})=8

\implies a(3-\sqrt{5})(3+\sqrt{5})=8

\implies4a=8

\implies a=2

Therefore, the equation of the parabola in factored form is:

\implies y=2(x-1-\sqrt{5})(x-1+\sqrt{5})

Expand so that the equation is in standard form:

\implies y=2(x^2-x+\sqrt{5}x-x+1-\sqrt{5}-\sqrt{5}x+\sqrt{5}-5)

\implies y=2(x^2-x-x+\sqrt{5}x-\sqrt{5}x+\sqrt{5}-\sqrt{5}+1-5)

\implies y=2(x^2-2x-4)

\implies y=2x^2-4x-8

6 0
2 years ago
Square root of 1.9321
Ann [662]
The answer to square root of 1.9321 is - 1.39
6 0
4 years ago
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