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3 years ago

What Is the sum of 6/1 and 4/1

Mathematics

1 answer:

Amiraneli [1.4K]3 years ago

6 0

6/1 = 6
4/1 = 4
6 + 4 = 10
10 is your answer.

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-6=b/18 solve for B

aleksandr82 [10.1K]

We can multiply both sides by 18:

-108 = b

7 0

3 years ago

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A map has a scale of 1 cm : 9 km. If two cities are 2 centimeters apart on the map, what is the actual distance between the citi

Angelina_Jolie [31]

Answer:

18 kilometers

Step-by-step explanation:

If each 1 cm represents 9 kilometers.

How many would 2 cm represent?

$x = \frac{9 \times 2}{1} = 9 \times 2 = 18$

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2 years ago

What is the solution to 4 (1/2x + 7) =12

Volgvan

Answer:

-8

Step-by-step explanation:

Solve the brackets first

4/2x+28=12

then group the like terms

2x=12-28

2x/2=-16/2

x= -8

i hope this helps

3 0

3 years ago

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What is -2+-2= Is it 4 yes or no

SSSSS [86.1K]

Answer:

-4

Step-by-step explanation:

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3 years ago

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

Therefore, the total outcomes for drawing straight are:

$N_S\;=\;({}^{4}C_1)*({}^{4}C_1)^5\;=\;10240$

Thus, the probability of getting a straight hand is:

$P(b)\;=\;\frac{N_S}{N_T}\\\\P(b)\;=\;\frac{10240}{2598960}\\\\P(b)\;=\; 0.0039$

6 0

3 years ago