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rewona [7]
3 years ago
6

Ross chose a card at random from a deck of 52 cards. Ross replaced the card and then chose a second card. What is the probabilit

y that Ross chose a club and then the ace of spades?
Mathematics
1 answer:
DaniilM [7]3 years ago
8 0

Answer:

\frac{1}{208}

Step-by-step explanation:

We have been given that Ross chose a card at random from a deck of 52 cards. Ross replaced the card and then chose a second card.

Let us find probability of getting a club.

\text{ Probability of getting a club}=\frac{13}{52} =\frac{1}{4}

Now let us find probability of getting the ace of spades.

\text{ Probability of getting ace of spades}=\frac{1}{52}

Now let us multiply these to find probability of getting a club and ace of spades.      

\text{ Probability of getting a club and then ace of spades}=\frac{1}{4}*\frac{1}{52}=\frac{1}{208}

Therefore, probability of getting a club and then the ace of spades \frac{1}{208}.

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Answer:

$44,000

Step-by-step explanation:

The statement indicates that Sandra saves 9% of her salary and she saved $4,050 this year. That means that the 9% of her salary, x, is equal to $4,050. You can write the following:

x*0.09= 4,050

Now, you can isolate x to find her salary this year:

x=4,050/0.09= 45,000

Then, you have that this year salary was 1000 more than in her previous year. As you have found her salary for this year, you can subtract 1,000 from it to find her salary in the previous year:

45,000-1,000= 44,000

According to this, the answer is that her salary in the previous year was $44,000.

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The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first deter
VMariaS [17]

Answer:

He must survey 123 adults.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

Assume that a recent survey suggests that about 87​% of adults have heard of the brand.

This means that \pi = 0.87

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

How many adults must he survey in order to be 90​% confident that his estimate is within five percentage points of the true population​ percentage?

This is n for which M = 0.05. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.05 = 1.645\sqrt{\frac{0.87*0.13}{n}}

0.05\sqrt{n} = 1.645\sqrt{0.87*0.13}

\sqrt{n} = \frac{1.645\sqrt{0.87*0.13}}{0.05}

(\sqrt{n})^2 = (\frac{1.645\sqrt{0.87*0.13}}{0.05})^2

n = 122.4

Rounding up:

He must survey 123 adults.

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