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luda_lava
3 months ago
14

A land surveyor places two stakes 500 ft apart and creates a perpendicular to the line that connects these two stakes. He needs

to place a third stake 100 ft away that is equidistant to the two original stakes. To apply the Perpendicular Bisector Theorem, the land surveyor would need to identify
A. a line parallel to the line connecting the two stakes

B. a line that is parallel to the perpendicular line

C. the midpoint along the line connecting the two stakes

D. the midpoint along any line that is perpendicular to the line connecting the two stakes
Mathematics
2 answers:
Svet_ta [12.7K]3 months ago
8 0

Answer:
We will select the final option as the correct one.
Step-by-step explanation:
A land surveyor positions two stakes 500 ft apart and identifies the midpoint between them.
From that midpoint, he is tasked with placing another stake 100 ft away, maintaining equal distance to the two original stakes.  
To utilize the Perpendicular Bisector Theorem, the land surveyor must identify a line that is "perpendicular to the segment connecting the two stakes and passes through the midpoint of those stakes."
Thus, we will choose the last option as the correct answer.

PIT_PIT [12.4K]3 months ago
8 0

Response:

D

Detailed explanation:

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J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
babunello [11817]

The 99% confidence interval for the actual mean difference between average mail-order and internet purchase amounts falls within [$(-31.82), $12.02].

Step-by-step clarification:

We know a random sample of 16 mail-order sales receipts shows a mean sale amount of $74.50 and a standard deviation of $17.25.

For internet sales, a random sample of 9 receipts gives a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal value utilized for constructing a 99% confidence interval for the true mean difference is given by;

                      P.Q.  =  

 ~

where,

= sample mean of mail-order sales = $74.50 \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }t__n_1_+_n_2_-_2

= sample mean of internet sales = $84.40

\bar X_1 = standard deviation for mail-order sales = $17.25

\bar X_2 = standard deviation for internet sales = $21.25

s_1 = number of mail-order sales receipts = 16

= number of internet sales receipts = 9s_2

Furthermore,  

 =  n_1 = 18.74

n_2

The actual mean difference between average mail-order and internet purchases is denoted by (s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }

).

Thus, the 99% confidence interval for (\mu_1-\mu_2) is expressed as;

      = \mu_1-\mu_2 Here, the t critical value at the 0.5% significance level with 23 degrees of freedom is 2.807.           =

          = [$-31.82, $12.02](\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Therefore, the 99% confidence interval for the true mean difference between average mail-order and internet purchases is [$(-31.82), $12.02].

(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

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An investment of $7,650 earns interest at the rate of 5% and is compounded quarterly. What is the accumulated value of the inves
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The width of a rectangle is 61 centimeters more than the length. The perimeter is 406 centimeters. Find the length and the width
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Step \; 1: \; Assign \; Variables \; for \; the \; unknown \; that \; we \; need \; to \; find

Let \; x \; be \; length \; of \; the \; rectangle

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Step \; 3: \; Solve \; the \; equation \; by \\ undoing \; whatever \; is \; done \; x.\\\\2(x+61+x)=406\\Group \; and \; Combine \; like \; terms \; inside \; the \; parenthesis\\\\2(2x+61)=406\\Distribute \; 2 \; in \; the \; left \; side \; of \; the \; equation\\\\4x+122=406\\Subtract \; 122 \; on \; both \; sides\\\\4x+122-122=406-122\\Simplify \; on \; both \; sides\\\\4x=284\\Divide \; on \; both \; sides\\\\\frac{4x}{4}=\frac{284}{4}\\Simplify \; fractions \; on \; both \; sides\\\\x=71

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