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2 years ago

What is the value of 5w if w = 6

Mathematics

2 answers:

Alenkasestr [34]2 years ago

7 0

Answer:

Step-by-step explanation:

if w=6 then 5w is equal to 5 * w

so therefore

5*6=30

hope this helps!

Vladimir79 [104]2 years ago

4 0

Answer: 30

Step-by-step explanation: In this problem, we are asked to evaluate the following expression if w = 6.

Since we know that w = 6, we can evaluate 5w by substituting 6 in for w. In addition, it's important to understand that 5w means 5 times w.

So substituting 6 in for w, we have 5(6).

Notice that I used parentheses around the 6. It's always a good idea to use parentheses when substituting numbers in for variables.

So we have 5(6) which is 30.

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You are planning to buy a house for $800,000. City bank offers a 30 year loan at 4.9 % apr ( Annual percentage interest rate) if

masya89 [10]

Answer:

3396.65

Step-by-step explanation:

Let's start by cacluating the amount the bank is loaning us

800000*.8=640000

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let x= payment

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4 0

2 years ago

Sos ,, helpp please !!!

Serhud [2]

8) x represents the number of pounds of peanuts, and y represents the number of pounds of raisins

9)
Solve:
x+2y=9
-2y -2y
x=9-2y

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-18 -18
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8 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 = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$