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brilliants [131]

3 years ago

15

10 POINTS!!!! Fast!!! Nick has designed a diamond-shaped kite as shown below. The measures of some sides of the kite are marked

in the figure. Find the value of x. (JUSTIFY)
Show your work.. thanks

1 answer:

choli [55]3 years ago

5 0

Answer: 3.5 in

Step-by-step explanation

∡DOC=90°

DC=7

OC=5

Using Pytagora equation:

DO=√DC²-OC²

DO=√49-25

DO=2√6

Use Equation for area of triangle:

area of ΔDOC=a*Ha/2

DC=a=7

Ha=x---hight of ΔDOC

Area of ΔDOC=DO*OC/2

2√6*5/2=5√6

5√6=7*x/2

10√6=7*x

x=10/7√6

x=24.5/7

X=3.5 in

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11/15-3/5 in fraction

Ket [755]

Answer:

l.c.m of 15 and 5 is 15

next step is to equalize the denominator with the l.c.m value

11/15 is the same (same denominator)

3/5 will change to 9/15 (multiply numerator and denominator by 3 to equalize)

Lastly, you just need to subtract

11/15 - 9/15

2/15

8 0

3 years ago

I will give branlyest

Anon25 [30]

Answer:

15 3 over 100

$2 \frac{1}{2} + 4 \frac{1}{4} + 9 \frac{1}{4} = 15 \frac{3}{10}$

5 0

3 years ago

What two numbers multiply to -126 and add up to -65

Airida [17]

Xy = -126
x + y = -65

$xy = -126$
$\frac{xy}{x} = \frac{-126}{x}$
$y = \frac{-126}{x}$

$x + y = -65$
$x + \frac{-126}{x} = -65$
$\frac{x^{2}}{x} + \frac{-126}{x} = -65$
$\frac{x^{2} - 126}{x} = -65$
$-65x = x^{2} - 126$
$0 = x^{2} + 65x - 126$
$x = \frac{-(65) \+ \sqrt{(65)^{2} - 4(1)(-126)}}{2(1)}$
$x = \frac{-65 \± \sqrt{4225 + 504}}{2}$
$x = \frac{-65 \± \sqrt{4729}}{2}$
$x = \frac{-65 \± 68.8}{2}$
$x = \frac{-65 + 68.8}{2}$ $or$ $x = \frac{-65 - 68.8}{2}$
$x = \frac{3.8}{2}$ $or$ $x = \frac{-133.8}{2}$
$x = 1.9$ $or$ $x = -66.9$

x + y = -65
  1.9 + y = -65
- 1.9   - 1.9
y = -66.9
  (x, y) = (1.9, -66.9)

or

  x + y = -65
  -66.9 + y = -65
+ 66.9 + 66.9
y = 1.9
  (x, y) = (-66.9, 1.9)

The two numbers that add up to -65 and can multiply to -126 are the numbers -66.9 and 1.9.

3 0

4 years ago

Explain the circumstances for which the interquartile range is the preferred measure of dispersion. What is an advantage that th

KatRina [158]

Answer:

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

Step-by-step explanation:

Previous concepts

The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

$IQR= Q_3 -Q_1$

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Solution to the problem

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

8 0

3 years ago

What is the quotient? The problem has been started for you.

ad-work [718]

Answer:89.6 (with line)

7 0

4 years ago

Read 2 more answers