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Delicious77 [7]

3 years ago

12

AABC is similar to ADEF. Find x. perimeter of triangle DEF= 60

1 answer:

vovangra [49]3 years ago

5 0

Find AC using Pythagorean theorem
6^2 + 8^2 = AC^2
36 + 64 = AC^2
100 = AC^2
10 = AC

6+8+10 = 24

24/60 = 2/5

10/x = 2/5
50 = 2x
25 = x

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Write the slopes intercept form of the equation of the line through the given points.

Anika [276]

Slope intercept form
y=mx+b
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(y2-y1)/(x2-x1)

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8 0

3 years ago

150.60 divided by 5/6 is ?

Lunna [17]

150.60 divided by 5/6 is 180.72.

7 0

3 years ago

If the coordinates of A are (1, 1) and the midpoint of AB is

yuradex [85]

Answer:

The coordinates of B are (-5, -1).

Step-by-step explanation:

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5 0

3 years ago

Find the vertex of the parabola.

Gnesinka [82]

Answer:

The vertex is (3,4)

Step-by-step explanation:

f (x) = x^2 - 6 x + 13

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4 0

3 years ago

According to a certain website, wine critics generally use a wine-scoring scale to communicate their opinions on the relative qu

const2013 [10]

Answer:

$\bar x = 82$

$\sigma = 9.64$

Step-by-step explanation:

Given

$n = 12$

$87\ 91\ 86\ 82\ 72\ 91\ 60\ 77\ 80\ 79\ 83\ 96$

Solving (a); Point estimate of mean

To do this, we simply calculate the sample mean

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{87+ 91+ 86+ 82+72+91+60+77+80+79+83+96}{12}$

$\bar x = \frac{984}{12}$

$\bar x = 82$

Solving (b); Point estimate of standard deviation

To do this, we simply calculate the sample standard deviation

$\sigma = \sqrt{\frac{\sum(x-\bar x)^2}{n - 1}$

$\sigma = \sqrt\frac{(87-82)^2+ (91-82)^2+ ....+ (79-82)^2+ (83-82)^2+ (96-82)^2}{12-1}$

$\sigma = \sqrt\frac{1022}{11}$

$\sigma = \sqrt{92.91}$

$\sigma = 9.64$

<em>Note that: The sample mean and the sample standard deviation are the best point estimators for the mean and the standard deviation, respectively.</em>

<em>Hence, the need to solve for sample mean and sample standard deviation</em>

3 0

3 years ago