Can I get some up please?

A tool set is on sale for $424.15.The original price of tthe tool set was $499.00. What percent of the original price is the sal

Jet001 [13]

Answer:

Step-by-step explanation:

The sale price is out of the original price so do 424.15 / 499.00 to find what it is out of in decimal form. Multiply your answer by 100 to find the percentage.

So:

424.15 / 499.00 =0.85

0.85 x 100 = 85

So the answer would 85%.

Hope this helps!

3 0

2 years ago

A group of friends share 3 1/2 slices of cakes each friend gets half a slice of cake how many friends shared the slices of cake

Mariana [72]

The blanks should be:
2 halves
7 halves
7 friends

This is because 3 and 1/2 is the same as the improper fraction of 7/2.

6 0

2 years ago

Read 2 more answers

A corporate team-building event costs $19plus an additional $1 per attendee. How many attendees can there be, at most, if the bu

Mars2501 [29]

Answer:

There can be at most $12$ attendees in a corporate team-building event.

Step-by-step explanation:

Let $x$ denotes number of attendees in a corporate team-building event.

Fixed cost = $19

Cost charged per attendee = $1

Budget for the corporate team-building event = $31

Therefore,

$19+1(x)\leq 31\\19+x\leq 31\\x\leq 31-19\\x\leq 12$

So, there can be at most $12$ attendees in a corporate team-building event.

8 0

2 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately 2020.022

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$