Ask question

Ask question

All categories

3 years ago

6

In a classroom, there are 13 scholars. Each scholar has a total of $1.75 in each pocket. If every scholar has a total of two poc

kets, how much money is in the classroom in total

1 answer:

mylen [45]3 years ago

5 0

13 x 1.75 = 22.75

take 22.75 and multiply by 2

you should get 45.50

You might be interested in

Please I need some answers

timama [110]

A) -8
b) -27
c) 32
d) X

3 0

2 years ago

What is the graph of y=-(x-2)^3-5

natulia [17]

When x=0,
y=-(0-2)^3-5
y=3

When y=0,
(x-2)^3=5
x-2= +/- 5^1/3
x= +/- 5^1/3 +2
x= 3.70997 or x=0.29002

To find the turning point at y, substitute the information that you have found above.

y=-(3.70997-2)^3-5
y=-9.99994

Find the mid-point next as mid-point is also the turning point.

(3.70997+0.29002)÷2=1.999995

Hence, Turning point is (1.999995, 9.99994)

Plot a graph using turning point, y value and the 2-x values.

6 0

3 years ago

How do you simplify 9/8?

Mademuasel [1]

Answer:

1 1/8

Step-by-step explanation:

divide 9 and 8 and u get ur answer Hoped I helped im Eve btw. Have a great day and consider marking this brainliest if you do thank you in advanced!

4 0

2 years ago

Help me solve this equation!<br> 4x + 10 = -2y<br> Remember to use the y = mbx+b formula!

Black_prince [1.1K]

Y=-2x-5
You can divide by -2 on both sides of the equal sign getting the answer above.

5 0

3 years ago

In the game of blackjack played with one deck, a player is initially dealt 2 different cards from the 52 different cards in the

Volgvan

Answer:

$P =4.83\%$

Step-by-step explanation:

First we calculate the number of possible ways to select 2 cards an ace and a card of 10 points.

There are 4 ace in the deck

There are 16 cards of 10 points in the deck

To make this calculation we use the formula of combinations

$nCr=\frac{n!}{r!(n-r)!}$

Where n is the total number of letters and r are chosen from them

The number of ways to choose 1 As is:

$4C1 = 4$

The number of ways to choose a 10-point letter is:

$16C1 = 16$

Therefore, the number of ways to choose an Ace and a 10-point card is:

$4C1 * 16C1 = 4 * 16 = 64$

Now the number of ways to choose any 2 cards from a deck of 52 cards is:

$52C2 =\frac{52!}{2!(52-2)!}$

$52C2 = 1326$

Therefore, the probability of obtaining an "blackjack" is:

$P = \frac{4C1 * 16C1}{52C2}$

$P = \frac{64}{1326}$

$P = \frac{32}{663}$

$P =0.0483$

$P =4.83\%$

4 0

3 years ago

Read 2 more answers