All categories

3 years ago

What is the probability of picking an ace from a standard deck of 52 cards?

Mathematics

1 answer:

katrin [286]3 years ago

8 0

There are 4 aces, so out of 52 cards you could have a 1/13 chance to get an Ace. That is 4/52 simplified.

You might be interested in

I would love some help :)

RUDIKE [14]

4x+6=3x+10 so x=4, to find the measure of KJL, plug in x into the equation 4x+6. angle KJL=22°

5 0

3 years ago

Can someone plz help me with this one problem

crimeas [40]

Answer:

24 1/2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Will give brainliest

inna [77]

Answer:

Step-by-step explanation:

PRBO

5 0

3 years ago

What is the slope of the line that passes

Snowcat [4.5K]

Answer:the answer is 5/2 because if yo subtract y2 from y1 and x2 from y2 thats the answer you get.

But you might not get that answer straight away you need to simplify the initial value and keep going until you cant simplify without a remainder.

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

Ms. Rivera created two dot plots to compare the results of her recent science quiz for 1st period students compared to 7th perio

Lorico [155]

The dot plots is an illustration of charts and graphs

The average score of period 7 is greater than the average score of period 1

<h3>How to determine the period with the higher average?</h3>

From the dot plots in the complete question, we have the following frequency tables

<u>Period 1</u> <u>Period 7</u>

Score Students Score Students

55 1 65 1

70 1 75 3

75 6 80 5

80 6 85 6

85 2 90 3

90 3 95 2

95 1

The average is calculated using:

$\bar x = \frac{\sum fx}{\sum f}$

For period 1, we have:

$\bar x_1 = \frac{55 * 1 + 70 * 1 + 75 * 6 + 80 * 6 + 85 * 2 + 90 * 3 + 95 * 1}{1 + 1 + 6 + 6 + 2 + 3 + 1}$

Evaluate the sum

$\bar x_1 = \frac{1590}{20}$

Evaluate the quotient

$\bar x_1 = 79.5$

For period 7, we have:

$\bar x_7 = \frac{65 * 1 + 75 * 3 + 80 * 5 + 85 * 6 + 90 * 3 + 95 * 2}{1 + 3 + 5 + 6 + 3 + 2}$

Evaluate the sum

$\bar x_7 = \frac{1660}{20}$

Evaluate the quotient

$\bar x_7 = 83$

By comparison, 83 is greater than 79.5

Hence, the average score of period 7 is greater than the average score of period 1