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3 years ago

Solve for r.

Mathematics

1 answer:

Serhud [2]3 years ago

8 0

If I assume that p is supposed to be r, then:

r = 12.5

r = 25/2 is improper or 12 & 1/2 is proper

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Help please Show two different ways to factor -2x - 10

Arada [10]

-2(x+5) OR 2(-x-5) I hope that helps if not please let me know!

3 0

3 years ago

The probability that a student has a Visa card (event V) is .63. The probability that a student has a MasterCard (event M) is .1

Aleksandr-060686 [28]

Answer:

a) The probability that a student has either a Visa card or a MasterCard is 0.71.

b) V and M are not independent.

Step-by-step explanation:

Given : The probability that a student has a Visa card (event V) is 0.63. The probability that a student has a MasterCard (event M) is 0.11. The probability that a student has both cards is 0.03.

To find :

a) The probability that a student has either a Visa card or a MasterCard ?

b) In this problem, are V and M independent ?

Solution :

The probability that a student has a visa card(event V) is P(V)= 0.63

The probability that a student has a MasterCard (event M) is P(M)= 0.11

The probability that a student has both cards is $P(V \cap M)=0.03$

a) Probability that a student has either a Visa card or a Master Card is given by,

$P(V \cup M) = P(V) + P(M) - P(V\cap M)$

$P(V \cup M) = 0.63+ 0.11- 0.03$

$P(V \cup M) =0.74- 0.03$

$P(V \cup M) =0.71$

The probability that a student has either a Visa card or a MasterCard is 0.71.

b) Two events, A and B, are independent if $P(A\cap B)=P(A)P(B)$

For V and M to be independent the condition is satisfied,

$P(V\cap M)=P(V)P(M)$

Substitute the values,

$0.03=0.63\times 0.11$

$0.03\neq 0.0693$

So, V and M are not independent.

6 0

3 years ago

How do I do 8b(ii) ? Please help me thank you!

Mila [183]

Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)] \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2 Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2 Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2 Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.

7 0

3 years ago

PLEASE HELP! i would really appreciate it thank you

LiRa [457]

Answer:

Step-by-step explanation:

5 0

2 years ago

Anyone? Please I need help!!

JulsSmile [24]

Answer:

the correct choice is marked

Step-by-step explanation:

Let x represent the smaller number. Then the larger number is 8x, and the difference is ...

8x -x = 280

7x = 280 . . . . . simplify

x = 40 . . . . . . divide by 7

The larger number is 8x = 8(40) = 320.

_____

Additional comment

Effectively, we have solved for the multiplier (40) that gives the ratio with 320 on top and a difference between top and bottom of 280:

$\dfrac{8}{1}=\dfrac{8\cdot40}{1\cdot40}=\dfrac{320}{40}$

4 0

2 years ago