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3 years ago

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(a) use the pythagorean theorem to determine the length of the unknown side of the triangle, (b) determine the perimeter of the

triangle, and (c) determine the area of the triangle. the figure is not drawn to scale.
the length of the unknown side is ____

the perimeter of the triangle is ____

the area of the triangle is ___

1 answer:

ki77a [65]3 years ago

7 0

Answer/Step-by-step explanation:

a. Unknown side, b, using the Pythagorean theorem is solved as shown below.

$b^2 = c^2 - a^2$

$b^2 = 45^2 - 27^2$

$b^2 = 1,296$

$b = \sqrt{1,296}$

$b = 36$

Unknown side, b, = 36 km

b. Perimeter of the triangle = sum of all the sides of the ∆ = $36 + 45 + 27$

$perimeter = 108 km$

c. Area of triangle = ½*base*height

where,

Base = 36 km

Height = 27 km

$Area = \frac{1}{2}*36*27$

$Area = 18*27$

$Area = 486 km^2$

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Angle BAC measures 56°.

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27 solid iron spheres, each of radius 'xcm'ate melted to form a sphere with radius 'ycm',find the ratio x:y

r-ruslan [8.4K]

Answer:3:1

Step-by-step explanation:

Given

There are 27 solid spheres of radius x cm

The volume of a sphere is $\frac{4}{3}\pi r^3$

The volume of 27 old spheres is the same as the volume of a new sphere

The volume of the old spheres is

$V_o=\dfrac{4}{3}\pi x^3$

The volume of a new sphere is

$V_n=\dfrac{4}{3}\pi y^3$

Now, $V_o=V_n$

$\Rightarrow 27\times \dfrac{4}{3}\pi x^3=\dfrac{4}{3}\pi y^3\\\\\Rightarrow (\dfrac{x}{y})^3=27\\\\\Rightarrow \dfrac{x}{y}=3$

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3 years ago

Mr. and Mrs. Storey drove 3,200 miles in all during their vacation. Mr. Storey drove 3 times as much as Mrs. Storey. How many mi

baherus [9]

Mr. drove 3 times more then Mrs.. ok

so how far Mrs. drove will be X
and Mr. will be Y
but we know that Mr. drove 3 times more then Mrs therefor
Y= 3X

and the Sum = 3,200

3,200=X+Y
3,200= X+(3X)
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div both by 4
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now we know how far Mrs drove now to find Mr.
again "Mr. drove 3 times more then Mrs"
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3 years ago

Find the probability of getting a king then an eight. Assume the cards are not replaced?

Nastasia [14]

There are 4 king and 4 eight cards inside 52 one deck of cards. The case asks for a king card and then an eight card.
To draw the king card, you have 4/52 probability.
To draw the eight card, you have 4/51 probability ( since the king card not replaced).

The total probability would be: 4*4/ 52*51= 4/663

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3 years ago

Suppose that the data for analysis includes the attributeage. Theagevalues for the datatuples are (in increasing order) 13, 15,

Bas_tet [7]

Answer:

a) $\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96$

$Median = 25$

b) $Mode = 25, 35$

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

c) $Midrange = \frac{70+13}{3}=41.5$

d) $Q_1 = \frac{20+21}{2} =20.5$

$Q_3 =\frac{35+35}{2}=35$

e) Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

f) Figura attached.

g) When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Step-by-step explanation:

For this case w ehave the following dataset given:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

Part a

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96$

The median on this case since we have 27 observations and that represent an even number would be the 14 position in the dataset ordered and we got:

$Median = 25$

Part b

The mode is the most repeated value on the dataset on this case would be:

$Mode = 25, 35$

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

Part c

The midrange is defined as:

$Midrange = \frac{Max+Min}{2}$

And if we replace we got:

$Midrange = \frac{70+13}{3}=41.5$

Part d

For the first quartile we need to work with the first 14 observations

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25

And the Q1 would be the average between the position 7 and 8 from these values, and we got:

$Q_1 = \frac{20+21}{2} =20.5$

And for the third quartile Q3 we need to use the last 14 observations:

25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70

And the Q3 would be the average between the position 7 and 8 from these values, and we got:

$Q_3 =\frac{35+35}{2}=35$

Part e

The five number summary for this case are:

Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

Part f

For this case we can use the following R code:

> x<-c(13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70)

> boxplot(x,main="boxplot for the Data")

And the result is on the figure attached. We see that the dsitribution seems to be assymetric. Right skewed with the Median<Mean

Part g

When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

6 0

3 years ago