What is a yearly expense you should expect to pay for your car?

valina [46]

A is the right answer I think

4 0

3 years ago

An elevator is rated to safely hold ten 250 lb people. The elevator is loaded as follows: three 600 lbs cartons of books, Jim (1

iris [78.8K]

To do this, we will need to find how much the elevator can hold in all. Multiply 10 by 250 to get 2500. Since it can also hold 10% more, multiply 2500 and 0.10 to find out how much 10% is. You get 250. Add 2500 and 250 to get how much the elevator can safely hold. It would be 2750.

To find out how much more weight can be added, first find out how much weight is on the elevator collectively. (books) 600+ (Jim) 180+ (Gavin) 330 + (Deb) 250 + (Susan) 120 = 1480. Subtract this from the amount the elevator can hold, and you get how much more weight can be added. 2750-1480= 1270.

The elevator can hold 1270 more ibs.

6 0

4 years ago

Read 2 more answers

Brandon spends his afternoon picking apples from an orchard. He notices that if he groups the apples he picked by 5's or 6's, th

marysya [2.9K]

Hello there!

a - 15 = 28k
a = 28k + 15
n = 30a + 3 = 30(28k + 15) + 3 = 840k + 453.
The only number that would fit between 1000 and 2000 is 1293.

In short, your answer is 1293.

Hope This Helps You!
Good Luck :)

3 0

3 years ago

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

You are asked to help out in the merchandise department. You sell $750 worth of $30 math teaching kits. How many kits did you se

Alex Ar [27]

$750 / $30 = 25 maths teaching kids

$350 / $25 = 14 maths starter kits
if you wanted to see if it was right, you would just do 14*25 and you would get $350

hope I helped!!

7 0

3 years ago