4x+2y=18 2x+3y=15 solution of the system?

A store manager paid $95 for an item and set the selling price at $116.85. What was the percent markup?

Neporo4naja [7]

We have been given that a store manager paid $95 for an item and set the selling price at $116.85. We are asked to find the percent mark-up.

We will use percent increase formula to solve our given problem.

$\text{Percent increase}=\frac{\text{Final}-\text{Initial}}{\text{Initial}}\times 100\%$

$\text{Percent increase}=\frac{\$116.85-\$95}{\$95}\times 100\%$

$\text{Percent increase}=\frac{\$21.85}{\$95}\times 100\%$

$\text{Percent increase}=0.23\times 100\%$

$\text{Percent increase}=23\%$

Therefore, the mark-up will be 23% and option 'b' is the correct choice.

6 0

3 years ago

Determine whether

lora16 [44]

Answer:

(a) and (b) are not equivalent

(c) is equivalent

Step-by-step explanation:

Given

$\frac{25^m}{5}$

Required

Determine an equivalent or nonequivalent expression

$(a)\ 25^{m-1$

We have:

$25^{m-1$

Apply law of indices

$25^{m-1} = \frac{25^m}{25}$

This is not equivalent to $\frac{25^m}{5}$

$(b)\ 25^{2m - 1}$

We have:

$25^{2m - 1}$

Apply law of indices

$25^{2m - 1} = \frac{25^{2m}}{25}$

This is not equivalent to $\frac{25^m}{5}$

$(c)\ 5^{2m-1}$

We have:

$5^{2m-1}$

Apply law of indices

$5^{2m-1} = \frac{5^{2m}}{5^1}$

$5^{2m-1} = \frac{5^{2m}}{5}$

Evaluate the numerator

$5^{2m-1} = \frac{25m}{5}$

This is an equivalent expression

4 0

3 years ago

Estimate the following quantities without using a calculator. Then find a more precise result, using a calculator if necessary.

zubka84 [21]

Explanation:

a. It is so easy to double a number that no approximation is necessary.

31,000 × 200 = 3.1×10⁴ × 2×10² = 6.2×10⁶ exactly

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b. 6.2×10³ × 5.2×10⁶ ≈ 6×5×10⁹ = 3×10¹⁰ approximately

The approximation can be refined a bit by taking the ".2" into account:

6.2×10³ × 5.2×10⁶ ≈ (6×5 + .2×(6+5))×10⁹

= 32.2×10⁹ = 3.22×10¹⁰ approximately

Actual product: 3.224×10¹⁰.

For most purposes, the approximation is an adequate approximation, as it it within 10% of the actual value.

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c. 9×10⁶ -2.3×10⁴ ≈ 9×10⁶ approximately

A better approximation is to actually subtract an approximation of the smaller number:

9×10⁶ -2.3×10⁴ ≈ (9 - 0.02)×10⁶ ≈ 8.98×10⁶ approximately

The actual value uses all of the digits of the smaller number:

9×10⁶ -2.3×10⁴ ≈ (9 - 0.023)×10⁶ = 8.977×10⁶ exactly

As in part B, either approximation is adequate for most purposes, as the difference from the actual value is less than .3%.

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The accuracy required of an approximation, hence the work you expend improving accuracy, should depend on the need in the final application of the number. Often, approximations are used for budget or resource planning purposes where some "slop" is allowed or even expected.

They can also be used in engineering applications, where the error needs to be on the side of more safety (rather than less).

7 0

3 years ago

Solve for b: 3b – 5 = 43

Thepotemich [5.8K]

we have the following: