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3 years ago

Suppose the sales tax rate is 5% on a $1300 tv how much is is the amount of tax?

Mathematics

1 answer:

shepuryov [24]3 years ago

3 0

Answer:

Therefore, amount of tax is $65.

Step-by-step explanation:

Amount of tax = Tax% of $1300

Amount of tax = 5% of $1300

= (5/100) of $1300

= (1/20) of $1300

= $130/2

= $65

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Q-35<br><br> Convert this equation to slope-intercept form: y+2= -4/17x + 12/7

Andru [333]

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept)

given y + 2 = - $\frac{4}{17}$ x + $\frac{12}{7}$

Subtract 2 from both sides

y = - $\frac{4}{17}$ x + $\frac{12}{7}$ - $\frac{14}{7}$

y = - $\frac{4}{17}$ x - $\frac{2}{7}$ ← in slope- intercept form

6 0

3 years ago

Name the minor arc and find its measure.

CaHeK987 [17]

The problem ask to find and choose the correct choice in the following choices that states the name of the minor arc and its measure base on the given date in your diagram. The answer would be letter D. arc AB;174 degree. I hope you are satisfied with my answer

8 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

Suppose a city with population 600,000 has been growing at a rate of 3% per year. If this rate continues, find the population

andrezito [222]

Answer:

The answer should be around 1,149663 (Rounded off to the nearest whole number. Also if a population has a remainer you usually add it as 1. This is because these are people)

Step-by-step explanation:

First step involves listing the formulas being used:

600,000= population, 0.03= growth rate 22years= period

Therefore, we take the 600,000 and multiply by 1+0.03 ^22(to the power of 22 or period)

So, 600,000(1+0.03)^22

=1,149662.045 /1,149663

Hope this helps!

6 0

3 years ago

What percent of 180 is 45? A. 4.5% B. 18% C. 22% D. 25%

Ber [7]

Answer:

D. 25%

Step-by-step explanation:

45/180=0.25

0.25 into a percent is 25%

6 0

3 years ago

Read 2 more answers