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damaskus [11]
3 years ago
7

Evaluate the expression (q/p)+3p^2 if q=3.5 and p=1

Mathematics
1 answer:
Julli [10]3 years ago
4 0

Answer:

The answer is 6.5

Step-by-step explanation:

(3.5/1)+3(1)^2

3.5/1=3.5

3(1)^2=3

3.5+3=6.5

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xenn [34]

Answer:

ASA

Step-by-step explanation:

In the 2 triangles

∠J = ∠L

side OJ = side IL

∠O = ∠I

The triangles are congruent by the angle/side/angle (ASA ) postulate

5 0
3 years ago
FInd the circumference of a circle with a diameter of 6 yd. Use the value 3.14 for x and do not round your answers
Varvara68 [4.7K]

Answer:

C ≈ 18.85yd

Step-by-step explanation:

5 0
2 years ago
A music club has 35 drum players. If 25% of the total number of members in the club are drum players, what is the total number o
Pachacha [2.7K]

Answer:

140

Step-by-step explanation:

Since 25% is 1/4 of 100, that means that 35 is 1/4 of the total amount of members in the club. So, 35 times 4 is 140.

3 0
3 years ago
What is the value of n in the equation –1/2(2n + 4) + 6 = –9 + 4(2n + 1)?
ioda

Answer:

<h2>n = 1</h2>

Step-by-step explanation:

-\dfrac{1}{2}(2n-4)+6=-9+4(2n+1)\qquad\text{use the distributive property}\\\\\left(-\dfrac{1}{2}\right)(2n)+\left(-\dfrac{1}{2}\right)(-4)+6=-9+(4)(2n)+(4)(1)\\\\-n-2+6=-9+8n+4\qquad\text{combine like terms}\\\\-n+(-2+6)=8n+(-9+4)\\\\-n+4=8n-5\qquad\text{subtract 4 from both sides}\\\\-n=8n-9\qquad\text{subtract}\ 8n\ \text{from both sides}\\\\-9n=-9\qquad\text{divide both sides by (-9)}\\\\n=1

7 0
3 years ago
Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on
Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

  • If he is given with two kings.
  • If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by =  \frac{^{3}C_2 }{^{47}C_2}   {∵ one king is already there}

              =  \frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}           {∵ ^{n}C_r = \frac{n!}{r! \times (n-r)!} }

              =  \frac{3}{1081}  =  0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =  \frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}

                                                                   =  \frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}

                                                                   =  \frac{6}{1081}  =  0.0055

Now, probability that he ends up with a full house = \frac{3}{1081} + \frac{6}{1081}

                                                                                    =  \frac{9}{1081} = <u>0.0083</u>.

3 0
3 years ago
Read 2 more answers
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