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2 years ago

A fair coin is tossed 5 times in a row. The exact probability of the coin landing heads exactly 2 times is?

Mathematics

1 answer:

OLga [1]2 years ago

5 0

$|\Omega|=2^5=32\\|A|=10\\\\P(A)=\dfrac{10}{32}=\dfrac{5}{16}$

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can somebody help me with these 4 questions I will give brainliest to anybody who answers them correctly

labwork [276]

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2 years ago

You deposit $2,200 in a bank account. Find the balance after 4 years if the account

nirvana33 [79]

Answer:

Step-by-step explanation:

2200(1+0.0225/12) raised to the power of 12(4)

2200(1.001875) raised to the power of 12(4)

2204.125 raised to the power of 12(4)

2.98771099E160

or 2.98771099 multiplied by 10 to the 160th power

3 0

2 years ago

You might need:

Nastasia [14]

Answer:

11 + x

Step-by-step explanation:

A = LW

W = $\frac{A}{L}$

Plug in given information:

W = $\frac{121-x^{2} }{11-x}$

Factor the numerator:

W = $\frac{(11+x)(11-x)}{(11-x)}$

Cancel (11 - x):

W = 11 + x

5 0

2 years ago

Use the Venn diagram to calculate probabilities

Shkiper50 [21]

From the Venn diagram, we can gather that there are 35 total objects (6 in both A and B; 15 in A but not B; 10 in B but not A; and 4 in neither A nor B), and we have the probabilities

$\mathbb P(A\cap B)=\dfrac6{35}$

$\mathbb P(A)=\dfrac{15+6}{35}=\dfrac{21}{35}=\dfrac35$ (this is the answer)

$\mathbb P(B)=\dfrac{10+6}{35}=\dfrac{16}{35}$

By definition of conditional probability,

$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{\frac6{35}}{\frac{16}{35}}=\dfrac6{16}=\dfrac38$

$P(B\mid A)=\dfrac{P(B\cap A)}{P(A)}=\dfrac{\frac6{35}}{\frac{21}{35}}=\dfrac6{21}=\dfrac27$

8 0

2 years ago

Read 2 more answers

Consider this equation.

antiseptic1488 [7]

Answer:

D: -√15/7

Step-by-step explanation:

Hopefully this helps!

8 0

2 years ago