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bezimeni [28]
3 years ago
10

Marisol has 486 square inches of wrapping paper

Mathematics
2 answers:
AfilCa [17]3 years ago
6 0

Answer:

9 inches

Step-by-step explanation:

faltersainse [42]3 years ago
3 0

Answer:

9 in

Step-by-step explanation:

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The ground floor of a tower has a very high ceiling, and the residential floors above it each have a constant, smaller, height.
seraphim [82]

Answer:

3.2 height

Step-by-step explanation:

54-31.6=22.4

15-8=7

22.4/7=3.2

4 0
3 years ago
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
Select all statements that are true about the linear equation.
Nataly [62]
A. True (plug in 0 for x to confirm y=-1)
b. True
c. True
d. False (If you plug in 1 for x, y would equal -2, not 2.)
6 0
2 years ago
Can you guys help me out pls
Slav-nsk [51]
It leaves Chicago at 11:49 am and takes 2 hrs 26 min.
So in central time, that gets you to Denver at 2:15 PM.
Since mountain time is one hour behind central time, your answer would be 1:15 PM.
5 0
3 years ago
Translate the following statement into a mathematical equation:
steposvetlana [31]
The answer is the first choice, 3(w-4)=5.

3 times = 3
a number minus four = (w-4)
is five = 5

I hope this helps.
8 0
3 years ago
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