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3 years ago

How many times does 104 go into 21? What is the remainder?

Mathematics

1 answer:

taurus [48]3 years ago

6 0

104 divided by 21 equals 4 with a remainder of 20, because 21 times 4 equals 84, and 104 - 84= 20.

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One wet weekend, 0.86 inches of rain fell on saturday. the next day, 1.52 inches of rain fell. how many inches of rain fell over

nataly862011 [7]

0.86 + 1.52 = 2.38 inches of rain fell over the weekend

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3 years ago

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Find the area and the circumference of the circle. Round your answers to the nearest hundredth.

Alik [6]

Answer:

$area = 200.96 \: {units}^{2} \\ circumfrence = 50.24 \: units$

Step-by-step explanation:

The formula to find the area and circumference of a circle is:

$area = \pi {r}^{2} \\ circumfrence = 2\pi \: r$

*note that pi is 3.14* Therefore the area is:

$formula = \pi {r}^{2} \\ 3.14 \times {8}^{2} = \\ 3.14 \times 64 = 200.96$

Therefore the circumference is:

$formula = \: 2\pi \: r \\ (2)(3.14)(8) = \\ 3.14 \times 16 = 50.24$

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2 years ago

A family went out to dinner and the cost of the food was $80. If the tip given was 15%, what

barxatty [35]

It should be $92.00

80 • .15 = 12
so 15% of the total ($80) is 12.

add the 12 to the total 80+12= $92

8 0

3 years ago

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In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any

Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

$E \cap F = \emptyset,\quad E \cup F = \Omega$

where $\Omega$ is the whole sample space.

This implies that

$P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)$

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day $d_1$ .

The second footballer can be born on any other day, so he has 364 possible birthdays:

$d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}$

the probability for the first two footballers to be born on two different days is thus

$1 \cdot \dfrac{364}{365} = \dfrac{364}{365}$

Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

$d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}$

so, the probability for the first three footballers to be born on three different days is

$1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}$

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

$\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

So, the probability that at least two footballers are born on the same day is

$1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

since the two events are complementary.

8 0

3 years ago

Which problem can be represented by the equation 100x+50=710?

andrew-mc [135]

Answer: The first answer

A coin bank can hold 710 coins. There are 50 coins in the bank. A student puts in 100 coins per week. How many weeks until the bank is full?

Step-by-step explanation:

7 0

3 years ago