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mamaluj [8]
3 years ago
7

In the figure below, PR and PQ are tangent to the circle with center O. Given that OQ = 10 and OP = 26, find PR.​

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
6 0
<h3>Answer: PR = 24</h3>

Explanation:

OQ = 10 is the radius, and so is segment RO. Both are the same length as they are the radii of the same circle. Triangle ORP has a leg of RO = 10 and a hypotenuse of PO = 26. The unknown side is PR = x.

Use the pythagorean theorem. We can use this theorem because the tangent formed (at point R) creates a 90 degree angle.

a^2 + b^2 = c^2

(PR)^2 + (RO)^2 = (PO)^2

x^2 + 10^2 = 26^2

x^2 + 100 = 676

x^2 = 676 - 100

x^2 = 576

x = sqrt(576) ... apply square root

x = 24

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Lunna [17]
ANSWER

T.S.A=1960m^2

EXPLANATION

The diagram represents the net of a rectangular prism.

The dimensions are,

l=42m

w = 14m

and

h = 7m

The total surface area of a rectangular prism is given by:

T.S.A=2(lw + lh + wh)

We substitute the values to obtain,

T.S.A=2(42 \times 14 + 42 \times 7 + 14 \times 7)

T.S.A=2(980)m^2


T.S.A=1960m^2
3 0
3 years ago
-3.8x - 5.9x = 223.1
katrin2010 [14]

Answer:

x=-23

Step-by-step explanation:

-3.8x-5.9x=223.1

-9.7x=223.1

x=223.1/-9.7

x=-23

5 0
3 years ago
Read 2 more answers
A large but sparsely populated county has two small hospitals, one at the south end of the county and one at the north end. The
Readme [11.4K]

Answer:

Step-by-step explanation:

Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by

f(x,y) = P((X=x,Y=y) ,

and given that the random variables are independent the joint pmf isbgiven by:

f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)

(b) the required probability is given by considering X=0,1 and Y =0,1

f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }

= [0.1 +0.2 ] × [0.1 + 0.3]

= 0.3 × 0.4

= 0.12

Now we find

fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1

fX(x) = sum{[f(x,y)]}= 0.1+0.3

= 0.4

fY(y) = sum{[f(x.y)]} = 0.1 + 0.2

= 0.3

Since fY(y) × fX(x) = 0.4 × 0.3

= 0.12

Hence f(x,y) = fX(x) .fY(y), the events are independent

(c) the required event is give by: [P(X<=1, PY<=1)]

= P(X<=1) . P(Y<=1)

= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1

= 0.4 × 0.3

= 0.12

(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A

A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]

A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]

A = [0.09] + [ 0.09]

A = 0.18

Pls note the sum of the vertical column X=2 is 0.3

3 0
3 years ago
Determine which numbers could not be used to represent the probability of an event. Select all that apply. A. StartFraction 64 O
vova2212 [387]

Answer:

A. \frac{64}{25} ​, because probability values cannot be greater than 1.

C. -1.5, because probability values cannot be less than 0.

Step-by-step explanation:

Probability is the extent to which an event is likely to happen. It ranges from 0(impossible) to 1(certain). Probability values can be written in decimal form or in fractional form.

The following numbers could not be used to represent the probability of an event.

A. \frac{64}{25} ​, because probability values cannot be greater than 1.

C. -1.5, because probability values cannot be less than 0.

6 0
3 years ago
Cos (90-theta) • cosec90-theta) =tano. How?​
denis-greek [22]

Answer:

<u>______________________________________________________</u>

<u>TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-</u>

For any right angled triangle with one angle α ,

  • \cos (90 - \alpha  ) = \sin \alpha  or  \sin(90 - \alpha ) = \cos\alpha
  • cosec \: (90 - \alpha  ) = \sec\alpha   or  \sec(90 - \alpha ) = cosec\:\alpha

<u>SOME GENERAL TRIGNOMETRIC FORMULAS :-</u>

  • <u></u>\sin \alpha = \frac{1}{cosec \: \alpha }  or  cosec \: \alpha  = \frac{1}{\sin \alpha }
  • <u></u>\cos \alpha = \frac{1}{\sec \alpha }  or  \sec \alpha = \frac{1}{\cos \alpha }

<u>______________________________________________________</u>

Now , lets come to the question.

In a right angled triangle , let one angle be α (in place of theta) .

So , lets solve L.H.S.

\cos (90 - \alpha ) \times cosec(90 - \alpha )

=> sin\alpha  \times \sec\alpha

=> \sin\alpha  \times \frac{1}{\cos\alpha }

=> \frac{\sin\alpha }{\cos\alpha }

=> \tan\alpha = R.H.S.

∴ L.H.S. = R.H.S. (Proved)

3 0
3 years ago
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