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2 years ago

Write 0.000073 in scientific notation

Mathematics

2 answers:

Llana [10]2 years ago

7 0

Answer:

7.3x10^-7

Step-by-step explanation:

you move the decimal to the right 7 times.

Hope this helps.

Andrew [12]2 years ago

7 0

Answer:

$10^-^5$ $*73$

Step-by-step explanation:

So in 0.000073, 7 is the fifth digit in the term so we can write it as $10^-^5$ * 73

or 10 to power of negative 5 multiplied by 73.

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Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

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Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.

$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

 $P(D)= \frac{1}{3}$ 

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

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Answer:

Step-by-step explanation:

a) this involves 2 tails

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Looking at 0.995 with df 60

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Answer:

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