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3 years ago

How many cm in 1.5 km

Mathematics

1 answer:

nignag [31]3 years ago

3 0

Answer:

1.5 Kilometers and 150000 Centimeters

Step-by-step explanation:

5 × 10 -5 kilometers in 1 1 / 2 centimeters. To convert any value in centimeters to kilometers, just multiply the value in centimeters by the conversion factor 1 / 100000 . So, 1.5 centimeters times 1 / 100000 is equal to 1.5 × 10 -5 kilometers.

I hope this helps!

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Liquid Tylenol has 325 milligrams of acetaminophen per 12.5 milliliters of liquid medicine (a fact that would be commonly abbrev

gulaghasi [49]

Answer:

10 milliliters

Step-by-step explanation:

260 mg/ 325 mg = x ml / 12.5 ml

12.5 ml x (260 mg / 325 mg) = 10 ml

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2 years ago

Mr. Chang’s class has 28 students, 4/7 of whom are girls. Which equation shows how to determine the number of girls in Mr. Chang

DedPeter [7]

Okay, so we know a number of students and the amount of girls as a bracket. So what we have to do now is multiply them with each other.
28 *4/7 = 16
So there are 16 girls in Mr. Chang´s class.
You can also determine the boys by - the students with the girls:
28-16=12 boys

Have a nice day :D

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3 years ago

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A math book comprises of 2 sections: arithmetic and geometry in the ratio 5:3. How much of each type of content will be needed t

Alex_Xolod [135]

480/8=60

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3 years ago

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2x + 11 + 6x<br> Combing like terms

Nataly [62]

Answer:8x+11

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add 2x to 6x and add 11 to it .

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2 years ago

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The mean annual income for adult women in one city is $28,520 and the standard deviation of the incomes is $5600. The distributi

pshichka [43]

Answer:

$E(\bar{x}) = 28520$

Standard error of mean = 689

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $28,520

Standard Deviation, σ = $5600

Mean of sampling distribution =

$E(\bar{x}) = \mu = 28520$

As per Central Limit Theorem, if the sample size is large enough, then the sampling distribution of the sample means follow approximately a normal distribution.

Sample size, n = 66

Since the sample size is large, we can use normal distribution for approximation.

Standard error of mean =

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{5600}{\sqrt{66}} = 689.31 \approx 689$

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2 years ago