Ask question

Ask question

All categories

3 years ago

14

A card was selected at random from a standard deck of cards. The suit of the card was recorded, and then the card was put back i

n the deck. The table shows the results after 40 trials.
What is the relative frequency of selecting a club?

17%

20%

25%

30%

PLEASE ANSWER QUICKLY

1 answer:

aleksklad [387]3 years ago

6 0

Answer:

30%

Step-by-step explanation:

You might be interested in

18% of what number is 46.8?

Rasek [7]

The answer is 260 have a good day.
I love it when people ask me these questions.

8 0

3 years ago

Read 2 more answers

alya collected 3 1/5 pounds of newspaper for the recycling Drive Rudy collected 4 1/5 times as much newspaper as a liar how many

iris [78.8K]

Answer:

13.44 pounds

Step-by-step explanation:

3.2x4.2=13.44 or 13 and 11/25

8 0

3 years ago

Find the fifth term in the sequence that is defined as follows a1=1. An=3an-1+2

faltersainse [42]

The fifth term in the sequence would be 161.

Step-by-step explanation:

We have to find the fifth term in the sequence $an=3a(n-1)+2$

Given that

a1 = 1 (that is the first term)

If we follow the formula, the value of every term depends on the value of immediate lower term. Therefore, the a2 can be calculated by putting 2 in place of n in above equation

$a2=3a(2-1)+2$

$a2=3a1+2$

$a2=3(1)+2$ as a1 = 1

$a2=5$

Similarly,

$a3=3a(3-1)+2$

$a3=3a2+2$

$a3=3(5)+2$ as a2 = 5

$a3=17$

and

$a4=3a(4-1)+2$

$a4=3a3+2$

$a4=3(17)+2$ as a3= 17

$a4=53$

and

$a5=3a(5-1)+2$

$a5=3a4+2$

$a5=3(53)+2$ as a4 = 53

$a5=161$

Therefore, the fifth term in the sequence would be 161.

Learn more:

The following links have more information about sequencing

brainly.com/question/7221312

Keywords: sequence, arithmetic series

#learnwithBrainly

6 0

4 years ago

SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

Inga [223]

Remember to be careful with signs when adding and subtracting. -g^4-10g^2h^2-3h^4

8 0

3 years ago

Alice searches for her term paper in her filing cabinet, which has several drawers. She knows thatshe left her term paper in dra

katen-ka-za [31]

You made a mistake with the probability $p_{j}$ , which should be $p_{i}$ in the last expression, so to be clear I will state the expression again.

So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer $j$ , is given by:

(1) $\frac{p_{j} }{1-d_{i}p_{i} } , if j \neq i,$ and

(2) $\frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} } , if j = i.$

so we can say:

$A$ is the event that you search drawer $i$ and find nothing,

$B$ is the event that you search drawer $i$ and find the paper,

$C_{k}$ is the event that the paper is in drawer $k, k = 1, ..., n.$

this gives us:

$P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}$

$P(A) = 1 - P(B) = 1 - d_{i} p_{i}$

Solution to Part (1):

if $j \neq i$ , then $P(A \cap C_{j} ) = P(C_{j} )$ ,

this means that

$P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)} = \frac{P(C_{j} )}{P(A)} = \frac{p_{j} }{1-d_{i}p_{i} }$

as needed so part one is solved.

Solution to Part(2):

so we have now that if $j$ = $i$ , we get that:

$P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}$

remember that:

$P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}$

this implies that:

$P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )$

so we just need to combine the above relations to get:

$P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} }$

as needed so part two is solved.

8 0

4 years ago