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2 years ago

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What is 0.02349 written in standard form to 3 significant figures

1 answer:

zalisa [80]2 years ago

4 0

Solve from right to left

$\\ \sf\Rrightarrow 0.02349$

$\\ \sf\Rrightarrow 0.0235\times 10^{-1}$

$\\ \sf\Rrightarrow 2.35\times 10^{-3}$

Option C is correct

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ExtremeBDS [4]

Answer:

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What is the name of the property of multiplication shown the equation.

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C- Commutative property.

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Monique buys a $4,700 air conditioning system using an installment plan that requires 15% down. How much is the down payment?

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3 years ago

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If there are approximately 1.8 million cars in Vancouver and if each car travels an average of 17,500 kilometres in a year, how

Goryan [66]

Rates are used to measure a quantity over another.

<em>The 1.8 million cars use </em> $1.85 \times 10^{10}$ <em> liters each year</em>

Given

$Cars = 1.8\ million$

$d = 17500km$ --- distance

$r = 40miles/gallon$

First, we calculate the number (n) of gallons used by each car

$r = \frac{d}{n}$

Solve for n

$n = \frac{d}{r}$

So, we have:

$n = \frac{17500km}{40miles/gallon}$

$n = \frac{17500km}{40miles}gallon$

Convert miles to kilometers

$n = \frac{17500km}{40km \times 1.60934}gallon$

$n = \frac{17500km}{64.3738km}gallon$

$n = 271.85\ gallon$

The number of gallons (N), used by all the cars is:

$N = n \times cars$

$N = 271.85 \times 1.8\ million$

$N = 489330000$

Convert to liters

$N = 489330000 \times 3.78541L$

$N = 1852314675.3L$

In scientific notation to 2 decimal places, we have:

$N = 1.85 \times 10^{10}L$

<em>Hence, the number of liters used is </em> $1.85 \times 10^{10}$ <em />

Read more about distance and rates at:

brainly.com/question/24659604

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3 years ago