Which statement is true?

Mathematics

1 answer:

4vir4ik [10]4 years ago

4 0

The correct answer is D because you cannot have a fraction as the base of a log

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Let B1W2 denote the outcome that the first ball drawn is B1 and the second ball drawn is W2. Because the first ball is replaced

Sati [7]

Answer:

$\{B_1B_1, B_1B_2, B_1W_1, B_1W_2, B_1W_3, \\B_2B_1, B_2B_2, B_2W_1, B_2W_2, B_2W_3, \\$

$W_1B_1,W_1B_2, W_1W_1, W_1W_2, W_1W_3, \\W_2B_1, W_2B_2, W_2W_1,W_2W_2, W_2W_3, \\W_3B_1, W_3B_2, W_3W_1, W_3W_2, W_3W_3\}$

Step-by-step explanation:

$\text{If an urn contains two blue balls} (denoted \:B_1 \:and \:B_2) \text{and three white balls}$ , $(denoted \:W_1, W_2, \:and\: W_3)$

If One ball is drawn, its color is recorded, and it is replaced in the urn. Then another ball is drawn and its color is recorded.

The 25 Possible outcomes of this experiment are listed below:

$\{B_1B_1, B_1B_2, B_1W_1, B_1W_2, B_1W_3, \\B_2B_1, B_2B_2, B_2W_1, B_2W_2, B_2W_3, \\$

$W_1B_1,W_1B_2, W_1W_1, W_1W_2, W_1W_3, \\W_2B_1, W_2B_2, W_2W_1,W_2W_2, W_2W_3, \\W_3B_1, W_3B_2, W_3W_1, W_3W_2, W_3W_3\}$

The tree diagram of this event is also attached.

6 0

4 years ago

In the game of euchre, the deck consists of the 9, 10, jack, queen, king and ace of each suit. players are dealt a five card han

Oksana_A [137]

0% because the deck only has 9s, 10s, jacks, queens, kings, and aces. There is no others cards accept those. So you cannot get a 4 of hearts because it’s not in the deck.

8 0

3 years ago

Read 2 more answers

Find the probability and interpret the results. if convenient, use technology to find the probability. during a certain week th

salantis [7]

From the given question:

$\mu=\$2.715 \\ \\ \sigma=\$0.049$

The probability of a normally distributed data between two values (a, b) is given by:

$P(a\ \textless \ \bar{x}\ \textless \ b)=P\left( \frac{b-\mu}{\sigma/\sqrt{n}} \right)-P\left( \frac{a-\mu}{\sigma/\sqrt{n}} \right)$

Thus,

$P(\$2.691\ \textless \ \bar{x}\ \textless \ \$2.732)=P\left( \frac{\$2.732-\$2.715}{\$0.049/\sqrt{32}} \right)-P\left( \frac{\$2.691-\$2.715}{\$0.049/\sqrt{32}} \right) \\ \\ =P(1.963)-P(-2.771)=0.97515-0.0028=\bold{0.97235}$