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3 years ago

Need help please!!!Given parallelogram MNOP if PN=27 what is NQ

Mathematics

1 answer:

likoan [24]3 years ago

8 0

9514 1404 393

Answer:

NQ = 13.5

Step-by-step explanation:

The diagonals of a parallelogram bisect each other. NQ is half of NP, so is ...

NQ = PN/2 = 27/2

NQ = 13.5

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3 years ago

How long must a ladder be to reach the top of a 12-foot wall if the both of the ladder is placed 3 feet from the base

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Calculate the distance between (4,9) and (-2,6) using the distance formula.

MAVERICK [17]

Answer:

$\large\boxed{d=3\sqrt5}$

Step-by-step explanation:

$\text{the formula of a distance between two points}\ A(x_1,\ y_1)\ \text{and}\ B(x_2,\ y_2):\\\\|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\=========================\\\\\text{We have}\ (4,\ 9)\ \text{and}\ (-2,\ 6).\ \text{Substitute:}\\\\d=\sqrt{(-2-4)^2+(6-9)^2}=\sqrt{(-6)^2+(-3)^2}=\sqrt{36+9}=\sqrt{45}\\\\\sqrt{45}=\sqrt{9\cdot5}=\sqrt9\cdot\sqrt5=3\sqrt5$

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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: $({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]$