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3 years ago

Predicate patterns for 1,1,2,3,5,8,13,21

Mathematics

1 answer:

ki77a [65]3 years ago

5 0

Answer:

If you add the number to the left to the number on the right it adds up to the number next to the right number

Step-by-step explanation:

1,1,2,3,5,8,13,21

1+1 is 2

2+1 is 3

3+2 is 5 and so on- hope this makes sense lol

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What are the opposites of 7, -3.5 , 2.25, and 9 1/3

Ber [7]

The opposite of 7 is -7, the opposite of -3.5 is 3.5, the opposite of 2.25 is -2.25, and the opposite of 9 1/3 is -9 1/3

7 0

4 years ago

Can someone help me on this ;-;

goldfiish [28.3K]

Answer:

4m+4

4(m+1)

Step-by-step explanation:

2m+m+m +2+2

Combine like terms

4m+4

Factor out 4

4(m+1)

6 0

3 years ago

Read 2 more answers

Which is the length of the hypotenuse of the right triangle? Round your answer to the nearest tenth of a centimeter. Hint: Pytha

jonny [76]

Answer:

$c =\sqrt{a^{2}+b^{2} }$

Step-by-step explanation:

You clear c, wich is the hypotenuse

$c =\sqrt{a^{2}+b^{2} }$

8 0

4 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

Rudik [331]

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