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Setler79 [48]
3 years ago
7

You are dealt two cards without replacement what is the probability the first card is one of 4 twos and second is 4 tens in a de

ck
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
8 0
A = drawing a two card
B = drawing a ten card

P(A) = 4/52 = 1/13
P(B|A) = P(A and B)/P(A)
P(A)*P(B|A) = P(A and B)
P(A and B) = P(A)*P(B|A)
P(A and B) = (1/13)*(4/51)
P(A and B) = 4/663
P(A and B) = 0.006033

The answer as a fraction is exactly 4/663
The answer in decimal form is approximately 0.006033
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Here is a set of signed numbers: 7, -3, 1/2, -0.8, 0.8, -1/10, -2 Order the numbers from least to greatest. I put the order -1/1
vichka [17]

Answer:

You answer is incorrect. The correct answer is -3, -2, -0.8, -1/10, 1/2, 0.8, 7.

Step-by-step explanation:

-3 is the smallest and 7 is the greatest.

1/2 is 0.5, which is smaller than 0.8.

-1/10 is -0.1, which is greater than -0.8.

-2 is greater than -3.

8 0
3 years ago
Need help with math. It says to simplify the expression 7(7t+6r+2)
RSB [31]
7 times 7 is 49
7 times 6 is 42
and 7 times 2 is 14
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3 years ago
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A is your answer

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3 years ago
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You have a chess board as pictured, with squares numbered 1 through 64. You also have a huge change jar with an unlimited number
ANEK [815]

Answer:

(a) 512 dimes

(b)D(n) = 2^{n-1}

(c) 9.22337 x 10¹⁸ dimes

(d) 9.22337 x 10¹² km

(e) roughly 61,489.13 times longer than the distance from the earth to the sun.

Step-by-step explanation:

b. Starting with 1 dime in the first square, if the number of dimes per each square is the double of the previous square, the general equation for the number of dimes in each square 'n' can be found by:

D(1) = 1\\D(2) = 2^1\\D(3) = 2^2\\D(4) = 2^3\\D(n) = 2^{n-1}

a. On the 10th square:

D(10) = 2^{10-1}\\D(10) = 512\ dimes

c. On the 64th square:

D(64) = 2^{64-1}\\D(10) = 9.22337 *10^{18}\ dimes

d. If a dime is 1 mm thick, the 64th pile will be:

h= 9.22337*10^{18}*0.0001\ m\\h= 9.22337*10^{15}\ m\\h= 9.22337*10^{12}\ km

e. The ratio between the height of the last pile (h) to the distance from the earth to the sun is:

R=\frac{9.22337*10^{12}\ km}{150,000,000\ km}\\ R=61,489.13

The height of the last pile is roughly 61,489.13 times longer than the distance from the earth to the sun.

5 0
3 years ago
I need help on this problem plz
Alborosie

Answer:

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Step-by-step explanation:

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