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MAXImum [283]
2 years ago
13

There are federal excise taxes on the retail price when purchasing fishing equipment. the taxes are intended to help pay for par

ks and conservation. what is the federal excise tax rate if there is an excise tax of $17.50 on a fishing rod and reel that has a retail price of $175?
Mathematics
1 answer:
Marizza181 [45]2 years ago
8 0

Using proportions, it is found that the federal tax rate is of 10%.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

The tax rate is the <u>proportion that $17.50 is of $175</u>, hence:

r = 17.50/175 = 0.1 = 10%.

The federal tax rate is of 10%.

More can be learned about proportions at brainly.com/question/24372153

#SPJ1

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What is y=4/5x+2 written in standard form
shepuryov [24]

Answer:

4x-5y=-10

Step-by-step explanation:

we know that

The equation of the line in standard form is in the form

Ax + By = C

where

A is a positive integer, and B, and C are integers

In this problem we have

y=\frac{4}{5}x+2 ----> equation of the line in slope intercept form

Convert to standard form

Multiply by 5 both sides

5y=4x+10

4x-5y=-10

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3 years ago
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Write a word phrase for c – 9.1. 9.1 less than a number a number subtracted from 9.1 a number plus 9.1 9.1 minus a number
xxTIMURxx [149]
<span>9.1 minus a number is correct</span>
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3 years ago
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Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

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

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
Write two expressions that are equivalent to 2(x+4)​
kotegsom [21]

(x+4)+(x+4)

2(x+2(2)) the (2) would be the exponent for 2 in the second one.

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3 years ago
A grocery store owner is planning a display by stacking cans of soup in a pyramid. The top row of the stack will have 1 can, the
Artyom0805 [142]

Answer:

144?

Step-by-step explanation:

1+3+5+7+9+11 blah blah blah...

1+3+5+7+9+11+13+15+17+19+21+23 which gives 144.

5 0
3 years ago
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