All categories

3 years ago

Hayden has 6 rolls of dimes. There are 50 dimes in each roll. How many dimes does he have altogether?

Mathematics

2 answers:

Darya [45]3 years ago

4 0

6×50=300
He has 300 dimes

Phoenix [80]3 years ago

3 0

50x6=300

you would just multiply

You might be interested in

A car dealership calculates the total number of tires on the lot using the function t(c)=4c where c represents the number of car

xxTIMURxx [149]

The number of cars (c)

5 0

3 years ago

What is the value of x that makes the statement true? 9x - 22 = 11x - 54 x =

storchak [24]

Answer:

Step-by-step explanation:

9x-22=11x-54 take 9x from each side

-22=2x-54 add 54 to each side

32=2x divide by 2

16=x

4 0

2 years ago

When you graph the exact same equation twice ?

Helen [10]

You will have infinite solutions

3 0

3 years ago

Find the probability of getting four consecutive aces when four cards are drawn without replacement from a standard deck of 52 p

posledela

Answer:

P=0.0000037

P=0.00037%

Step-by-step explanation:

Probability

A standard deck of 52 playing cards has 4 aces.

The probability of getting one of those aces is

$\displaystyle \frac{4}{52}=\frac{1}{13}$

Now we got an ace, there are 3 more aces out of 51 cards.

The probability of getting one of those aces is

$\displaystyle \frac{3}{51}=\frac{1}{17}$

Now we have 2 aces out of 50 cards.

The probability of getting one of those aces is

$\displaystyle \frac{2}{50}=\frac{1}{25}$

Finally, the probability of getting the remaining ace out of the 49 cards is:

$\displaystyle \frac{1}{49}$

The probability of getting the four consecutive aces is the product of the above-calculated probabilities:

$\displaystyle P= \frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{27}\cdot\frac{1}{49}$

$\displaystyle P= \frac{1}{270,725}$

P=0.0000037

P=0.00037%

3 0

2 years ago

Oliver works at a bookstore.He packed 20 identical paperbacks and 9 identical textbooks in a box. The total mass of the books wa

olchik [2.2K]

If we let p and t be the masses of the paper and textbook, respectively, the equations that would best represent the given in this item are:
(1) 20p + 9t = 44.4
(2) (20 + 5)p + (9 + 1)t = 51

The values of p and t from the equation are 0.6 and 3.6, respectively. Thus, each paperback weighs 0.6 pounds and each textbook weighs 3.6 pounds.

7 0

3 years ago