Need help on this please

Two cards are randomly selected from an ordinary playing deck.

Anna007 [38]

Answer:

The required probability is 0.048

Step-by-step explanation:

Consider the provided information.

Out of 52 card we need to select only 2.

Therefore the sample space is: $^{52}C_2=1326$

One card is Ace. The number of Ace are 4 and we need to select one of them.

This can be written as: $^4C_1=4$

The second card can be Ten or a Jack or a Queen or a King.

There are 4 Ten, 4 jack, 4 Queen, and 4 king in a deck and we need to select only 1 out of them.

So we can say that we need to select 1 card out of 16.

This can be written as: $^{16}C_1=16$

Thus, the probability that one of the cards is an ace and the other one is either a ten, a jack, a queen or a king is:

$P=\frac{16\times4}{1326}\approx0.048$

Hence, the required probability is 0.048

3 0

4 years ago

7th grade math help me please :)))

yanalaym [24]

Answer:

$284.99 should go to Mr. Aceves’ class, $174.41 should go to Mrs. Baca’s class, and $140.60 should go to Mr. Canyon’s class.

Step-by-step explanation:

Firstly, to calculate the portions each class should have, we should find the proportion each class takes up.

Hence, let's say A represents Mr. Aceves' class, B represents Mrs. Baca's class, and C represents Mr. Canyon's class.

The total number of box tops collected would be 3760 + 2301 + 1855 = 7916 box tops.

Therefore, the total proportions would be:

A = $\frac{3760}{7916}$

B = $\frac{2301}{7916}$

C = $\frac{1855}{7916}$

Hence, we can finally multiply these proportions with the total prize to find the appropriate division:

Mr. Aceves' class gets $\frac{3760}{7916} \cdot 600 \approx 284.99$ .

Mrs. Baca's class gets $\frac{2301}{7916} \cdot 600 \approx 174.41$ .

And Mr. Canyon's class gets $\frac{1855}{7916} \cdot 600 \approx 140.60$ .

Finally, $284.99 should go to Mr. Aceves’ class, $174.41 should go to Mrs. Baca’s class, and $140.60 should go to Mr. Canyon’s class.

Hope this helped!

7 0

3 years ago

HELP HELP PLEASSEEE I HAV 2 MINUTES LEF

Brums [2.3K]

Answer:

A because I have had this question before

Step-by-step explanation:

trust

6 0

3 years ago

Read 2 more answers

How much can a 12cm high

QveST [7]

12 cm high is about 4.7 inches high

3 0

3 years ago

Read 2 more answers

7. Which of the following can be determined from the graph below?

tatyana61 [14]

Answer:

The ordered pair (6,25) is a solution to both

$y=-\frac{5}{2}x+40$ and $y=\frac{5}{3}x+15$

Step-by-step explanation:

Part 7)

step 1

Find the equation of the line with positive slope

take the points (0,15) and (9,30)

Find the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the given values

$m=\frac{30-15}{9-0}$

$m=\frac{15}{9}$

simplify

$m=\frac{5}{3}$

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$m=\frac{5}{3}$

$b=15$

substitute

$y=\frac{5}{3}x+15$

step 2

Find the equation of the line with negative slope

take the points (0,40) and (8,20)

Find the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the given values

$m=\frac{20-40}{8-0}$

$m=-\frac{20}{8}$

simplify

$m=-\frac{5}{2}$

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$m=-\frac{5}{2}$

$b=40$

substitute

$y=-\frac{5}{2}x+40$

step 3

Find the solution of the system

we know that

The solution of the system of equations is the intersection point both graphs

The intersection point is (6,25) ----> see the graph

therefore

The ordered pair (6,25) is a solution to both $y=-\frac{5}{2}x+40$ and

$y=\frac{5}{3}x+15$

7 0

3 years ago