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dedylja [7]
3 years ago
9

A standard deck of playing cards has 52 cards: 13 spades, 13 clubs, 13 hearts, and 13 diamonds. What is the probability of drawi

ng a spade from a standard 52-card deck, replacing it, and then drawing another spade?
Mathematics
1 answer:
Vaselesa [24]3 years ago
5 0

13 spades, 52 cards total

probability of drawing a single spade = 13/52 = 1/4

If we replace that card, then the probability of drawing of drawing a spade is still 1/4 as nothing has changed

Overall, the probability of getting two spades in a row is (1/4)*(1/4) = 1/16

note: the two events are independent because of the replacement happening

---------

The answer as a fraction is 1/16

As a decimal, that would be 0.0625 which converts to 6.25%

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Answer:

where are the options

Step-by-step explanation:

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The circumference of a circle is 257 cm. What is the area,<br> in<br> square<br> centimeters
nikdorinn [45]

Answer:

2\pi \times r = c

\pi \times  {r}^{2}  = a

A=5257.76

Step-by-step explanation:

Or use a calculator online.

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Based on the ratios shown ion the double number line, what is the cost for 5 pounds of almonds? :)
denis-greek [22]
I found the answer to this by first dividing 34 by 4 to get 8.5. We already know what 4 divided by 4 is. For every one pound of almonds, it costs $8.50. Since we know that we need to find out how much it costs for 5 pounds of almonds, we simply multiply 8.5 by 5 to get 42.5. It costs $42.50 to buy 5 pounds of almonds.


7 0
3 years ago
Suppose z equals f (x comma y ), where x (u comma v )space equals space 2 u plus space v squared, y (u comma v )space equals spa
barxatty [35]

z=f(x(u,v),y(u,v)),\begin{cases}x(u,v)=2u+v^2\\y(u,v)=3u-v\end{cases}

We're given that f_x(6,1)=3 and f_y(6,1)=-1, and want to find \frac{\partial z}{\partial v}(1,2).

By the chain rule, we have

\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}

and

\dfrac{\partial x}{\partial v}=2v

\dfrac{\partial y}{\partial v}=-1

Then

\dfrac{\partial z}{\partial v}(1,2)=\dfrac{\partial z}{\partial x}(6,1)\dfrac{\partial x}{\partial v}(1,2)+\dfrac{\partial z}{\partial y}(6,1)\dfrac{\partial y}{\partial v}(1,2)

(because the point (x,y)=(6,1) corresponds to (u,v)=(1,2))

\implies\dfrac{\partial z}{\partial v}(1,2)=3\cdot2\cdot2+(-1)\cdot(-1)=\boxed{13}

4 0
3 years ago
Seventh grade &gt; EE.11 Probability of independent and dependent events NED
ella [17]

Step-by-step explanation:

each combination of 2 specific numbers has the same probability. there is no difference in probability between the individual numbers.

remember, a probability is always desired cases over all possible cases.

we have 4 different possible outcomes every time we spin the spinner.

to get a specific number has the probability of 1/4, because we want 1 specific outcome, and have in total 4 different possibilities.

now, we spin a second time. the probability to get a specific number is again 1/4.

but, if we consider both events to be connected, when we want to know the probability to get 2 specific numbers when spinning twice, we have to multiply the individual probabilities :

1/4 × 1/4 = 1/16

so, the probability to land first on a 5, and then secondly on a 2 is 1/16.

the same as for landing first on a 3, and then on a 5.

the same as for landing first on a 4, and then on a 4 (again).

that is because the individual spin results are independent. the result of the first spin does not impact in any way the result of the second spin (in contrary to e.g. pulling multiple cards without returning the previously pulled cards).

8 0
2 years ago
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