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2 years ago

12

Consider a standard deck of 52 cards (13 cards each in 4 suits, clubs, spades, diamonds and hearts). What is the probability of

picking out 5 cards in a row which are all diamonds, without replacing any card to the deck?

1 answer:

Goryan [66]2 years ago

8 0

5/52 as you are taking 5 cards out

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Susie scored 85, 84,and 78 on three of the four math tests. if her average score for the four tests was 86. what was her score o

nata0808 [166]

(85 + 84 + 78 + x) / 4 = 86
(247 + x) / 4 = 86
247 + x = 86 * 4
247 + x = 344
x = 344 - 247
x = 97 <=== her score was 97

7 0

3 years ago

Read 2 more answers

- 1 13 and its opposite on a number line.

timofeeve [1]

Answer:

-113 and 113

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Determine what type of model best fits the given situation:

lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P = $\frac{20}{100} \times P = 0.2P$

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = $\frac{20}{100} \times 0.8P = (0.2)(0.8P)$

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

$=P(0.8)^{2}$ -------------------------(2)

Similarly after 3 years, this value $P(0.8)^{2}$ will again be decreased by 20 %

so 20% of $P(0.8)^{2} \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}$

So after 3 years value is decreased by $(0.2)P(0.8)^{2}$

so value after 3 years will be = $P(0.8)^{2} - (0.2)P(0.8)^{2}$

taking $P(0.8)^{2}$ common out we get $P(0.8)^{2}(1-0.2)$

$P(0.8)^{2}(0.8)$

$P(0.8)^{3}$ -----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or $P(0.8)^{1}$

value after 2 years is $P(0.8)^{2}$

value after 3 years is $P(0.8)^{3}$

so value after x years will be $P(0.8)^{x}$ ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

$y = P(0.8)^{x}$ where y is the value after x years

so thats the final answer

3 0

4 years ago

An aquarium tank in the shape of a rectangular prism is 56 cm long, 29 cm wide, and 24 cm high. The top of the tank is open, and

kobusy [5.1K]

0 if you add the thickness to the underside of the tank

5 0

3 years ago

Jan needed to empty 18 gallons of water from a tub into jugs that hold 0.5 gallons of water. How many jugs did Jan need?

Ilia_Sergeevich [38]

So you would do,
18 gallons of water divided by 0.5 gallons of water
if you did that correctly you should have gotten... 36 jugs

8 0

3 years ago