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3 years ago

Noah bought 2.5 kilograms of colored tile chips for a mosaic art project. What is the mass of the tile chips in grams?

Mathematics

1 answer:

fgiga [73]3 years ago

8 0

It is 2,500 grams because one kilogram equals to 1,000 grams

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Breanna has quarters, dimes, and nickels in her purse. she has 3 fewer nickles than dimes, but she has 2 more nickles than quart

zmey [24]

We base ourselves in the one piece of crucial evidence: She has 2 quarters.
-from the rest of the problem, we know that she has 2 more NICKEL'S than QUARTERS,

(2 Quarters + 2 = 4 nickels)

she has 3 fewer nickels than dimes,

DIME - 3 = Nickels

D - 3 = 4

D = 7

7 dimes
(70 cents)

4 nickels
(20 cents)

2 quarters
(25 cents)

115 cents
or $1.15

8 0

4 years ago

Read 2 more answers

Tasha invests $5000 annually at 6% and an additional $5000 annually at 8%. Thomas invests $10000 annually at 7%. Which statement

REY [17]

Answer:

Answer:

Option B is correct.

Step-by-step explanation:

We will compare the interest earned by both.

Tasha: p = $5000

r = 6% or 0.06

n = 1

So, Amount after a year will be = = $5300

And amount the next year with p = 5300: 5300*1.06= $5618

Additional $5000 at 8%

Here the amount will be = =5400

Next year amount with p = 5400 : 5400*1.08 = $ 5832

Amount in total Tasha will have in 2 years = 5618+5832 = 11450

Thomas:

p = 10000

r = 7% or 0.07

n = 1

After a year the amount will be = =$10700

Amount Next year with p = 10700 : 10700*1.07 = $11449

*****Just after 1 year we can see that Tasha's total amount is high than Thomas. This means at the same consistent rate, each year Tasha's amount will always be higher than Thomas.

So, option B is correct. Tasha’s investment will yield more over many years because the amount invested at 8% causes the overall total to increase faster.

6 0

3 years ago

Read 2 more answers

Can someone help me with my homework please

vichka [17]

Now you can see its pretty simple, its 128 but the way i got it is first you double 8 which makes 16 right then double 16 and you can see you got 32 then double 32 which makes 64 so next what you have to do is double 64 Obiously so you just double those numbers to get 128....Get it?

4 0

3 years ago

Please help me this is so confusing to me.

Naddik [55]

Hi there! :)

$\large\boxed{A = 120^{o}}$

Find the central angle using a ratio in regards to angle and circumference.

The radius is 27 units, so use the following equation to calculate circumference:

C = 2rπ

Therefore:

C = 2(27)π = 54π

Since we have the total circumference, we can write the following proportion:

$\frac{x}{360} = \frac{18\pi}{54\pi}$

Cross multiply:

54πx = 6480π

Divide both sides by 54π to isolate for x:

x = 120°

8 0

3 years ago

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

(c) The novels must be together and the chemistry books, together

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$