A building can be insured by multiple insurance companies at the same time. True or False?

What is the simple interest on $2500 invested at an interest rate of 6% for 2 years?

Brrunno [24]

Answer:

The simple interest is $300.

Step-by-step explanation:

<em>Find the Interest amount.</em>

I: Interest ?

P: Principle Amount $2500

R: Rate 6% = .06

T: Time Period (in years) 2 Years

<em>Convert into equation to find Simple Interest</em>

I = Prt

I = (2500)(.06)(2)

<em>Calculate</em>

I = 300

<em />

4 0

3 years ago

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The basketball team has 50 members, and 18 of them are 6th graders. What percentage of the team is in the 6th grade?

motikmotik

Answer:

36%

Step-by-step explanation:

$\frac{18}{50} = \frac{36}{100}$

So 36% of the team is in 6th grade.

8 0

3 years ago

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Let $s$ be a subset of $\{1, 2, 3, \dots, 100\}$, containing $50$ elements. how many such sets have the property that every pair

Tamiku [17]

Let A be the set {1, 2, 3, 4, 5, ...., 99, 100}.

The set of Odd numbers O = {1, 3, 5, 7, ...97, 99}, among these the odd primes are :

P={3, 5, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

we can count that n(O)=50 and n(P)=24.

Any prime number has a common factor >1 with only multiples of itself.

For example 41 has a common multiple >1 with 41*2=82, 41*3=123, which is out of the list and so on...

For example consider the prime 13, it has common multiples >1 with 26, 39, 52, 65, 78, 91, and 104... which is out of the list.

Similarly, for the smallest odd prime, 3, we see that we are soon out of the list:

3, 3*2=6, 3*3=9, ......3*33=99, 3*34=102..

we cannot include any non-multiple of 3 in a list containing 3. We cannot include for example 5, as the greatest common factor of 3 and 5 is 1.

This means that none of the odd numbers can be contained in the described subsets.

Now consider the remaining 26 odd numbers:

{1, 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}

which can be written in terms of their prime factors as:

{1, 3*3, 3*5, 3*7, 5*5,3*3*3, 3*11,5*7, 3*13, 2*2*3*3, 7*7, 3*17, 5*11 , 3*19,3*21, 5*13, 3*23,3*5*5, 7*11, 3*3*3*3, 5*17, 3*29, 7*13, 3*31, 5*19, 3*3*11}

1 certainly cannot be in the sets, as its common factor with any of the other numbers is 1.

3*3 has 3 as its least factor (except 1), so numbers with common factors greater than 1, must be multiples of 3. We already tried and found out that there cannot be produced enough such numbers within the set { 1, 2, 3, ...}

3*5: numbers with common factors >1, with 3*5 must be

either multiples of 3: 3, 3*2, 3*3, ...3*33 (32 of them)

either multiples of 5: 5, 5*2, ...5*20 (19 of them)

or of both : 15, 15*2, 15*3, 15*4, 15*5, 15*6 (6 of them)

we may ask "why not add the multiples of 3 and of 5", we have 32+19=51, which seems to work.

The reason is that some of these 32 and 19 are common, so we do not have 51, and more important, some of these numbers do not have a common factor >1:

for example: 3*33 and 5*20

so the largest number we can get is to count the multiples of the smallest factor, which is 3 in our case.

By this reasoning, it is clear that we cannot construct a set of 50 elements from {1, 2, 3, ....} containing any of the above odd numbers, such that the common factor of any 2 elements of this set is >1.

What is left, is the very first (and only) obvious set: {2, 4, 6, 8, ...., 48, 50}

<span>Answer: only 1: the set {2, 4, 6, …100}</span>

8 0

3 years ago

The swimming pool at Spring Valley High School is a rectangle with a width of 70 meters and a length of 25 meters. Around the pe

erik [133]

<h3>Answer: 4w^2+190w</h3>

================================================

Explanation:

Check out the attached image below. Specifically, take a look at figure 1. We start with a rectangle which is 70 meters by 25 meters. Then we extend this rectangle out by adding two copies of w to each dimension having the larger rectangle be (70+2w) meters by (25+2w) meters.

The original smaller rectangle has area of 70*25 = 1750 square meters.

The new larger rectangle area is a bit more tricky to figure out, though its not too bad once you get the hang of it. You could use the FOIL rule to get

(70+2w)(25+2w) = 70*25+70*2w+2w*25+2w*2w

(70+2w)(25+2w) = 1750+140w+50w+4w^2

(70+2w)(25+2w) = 1750+190w+4w^2

(70+2w)(25+2w) = 4w^2+190w+1750

Or you could use the box method shown in figure 2.

The idea is to write the terms of (70+2w) and (25+2w) along the edges of the box. Then filling out the 4 inner boxes involves multiplying the outer terms. Example: row1,column2 has 50w inside it because we multiply the outer terms 25 and 2w to get 25*2w = 50w. The other boxes are filled out in a similar fashion.

Although the FOIL rule is handy and taught in a lot of math books/classes, I prefer the box method because the box method can be extended to other cases (not just multiplying two binomials).

Whichever method you use, the larger rectangle has area 4w^2+190w+1750 square meters.

-------------------------------

So,

A = area of larger rectangle = 4w^2+190w+1750

B = area of smaller rectangle = 1750

C = difference in areas of A and B

C = A - B

C = ( A ) - ( B )

C = ( 4w^2+190w+1750 ) - ( 1750 )

C = 4w^2+190w+1750-1750

C = 4w^2+190w

This represents the area of just the tile border alone. We started with the larger rectangle and took out the smaller rectangle inside to only have the border left over.

-------------------------------

If you want, you can factor out a 'w' to get

4w^2+190w = w(4w+190)

Though this may be optional.

8 0

3 years ago

What is 1/3 of 20,000

kari74 [83]

6000 is the correct answer

4 0

2 years ago

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