What number has 2 more tens than 11?

1 answer:

3 0

Well I'd say it's any number before or after 11. Since it says 2 more tens it could be 31 by adding. You can also use algebra to help you.

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Write the equation in slope-intercept form of the line that contains (3,4) and is perpendicular to y=-2x-4

nataly862011 [7]

If a line is perpendicular to other his slopes are the opposite inverse of each other

So, if y = -2x - 4 we have the slope of -2, so the slope of the other line will be the opposite inverse which means

opposite of -2 = 2

inverse of 2 = 1/2

So our slope is 1/2

(y - yo) = m.(x - xo)

(y - 4) = 1/2.(x - 3)

y - 4 = x/2 - 3/2

y = x/2 - 3/2 + 4

y = x/2 + 5/2

5 0

3 years ago

Need a quick answer please!

ANTONII [103]

Answer:

It’s going to be b

Step-by-step explanation:

Because add then then divide then put the x on top and simplify then and then subtract

7 0

3 years ago

Five cards are dealt from a standard 52-card deck. (a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1

labwork [276]

Answer:

Step-by-step explanation:

As there are total 52 cards in a deck and we have to draw a set of 5 cards, we can use the formula of combination to find the total number of possible ways of drawing 5 cards.

Number of ways to draw 5 cards = $N_T$

$N_T\;=\;({}^NC_k)\\\\N_T\;=\;({}^{52}C_5)\\\\N_T\;=\;2,598,960$

(a) Assuming the cards are drawn in order (would not affect the probability). The of getting Ace, 2, 3, 4 and 5 can be obtained by multiplying the probability of getting cards below 6 (20/52) with the probability of getting 5 different cards (4 choices for each card).

$P(a)\;=\;\frac{20}{52}*\frac{4}{52}*\frac{4}{51}*\frac{4}{50}*\frac{4}{49}*\frac{4}{48}\\\\P(a)\;=\; 1.3133*10^{-6}$

(b) For a straight we require our set to be in a sequence. The choices for lowest value card to produce a sequence are ace, 2, 3, 4, 5, 6, 7, 8, 9, or 10. Hence, the number of ways are $({}^{10}C_1)$ .

For each card we can draw from any of the 4 sets. It can be described mathematically as: $({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]$