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The legend on a map states that 1 cm is 20 km. If you measure 9 centimeters on the map, how many kilometers would the actual distance be?

Answers

Answer 1
Answer:

Making a proportion helps solve this problem. Cm on top, km on bottom.


1 9

---- = ----

20 ?


Cross multiply to get 1? = 180


? = 180.


The distance is 180 km.
Answer 2
Answer:

Final answer:

The conversion of 9 cm on the map to an actual distance, using the given scale of 1 cm for 20 km, results in an actual distance of 180 kilometers.

Explanation:

The question is asking for the actual distance corresponding to 9 cm on the map. According to the given scale on the map, 1 cm corresponds to an actual distance of 20 km. So to find out how many kilometers 9 cm on the map would be in real life, we simply multiply the length measured on the map by the distance each centimeter represents. This gives us:

9 cm * 20 km/cm = 180 km

So, according to the map, 9 cm corresponds to an actual distance of 180 kilometers.

Learn more about Map Scale Conversion here:

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An object weighs 1 pound on earth weighs about 1/15 pounds on Pluto. If a man weighs 240 pounds on earth how many pounds would he weigh on Pluto?

Answers

Answer:

16 I believe. 240 divided by 15 is 16.

What is a rate in math?

Answers

Answer:

  A rate is a ratio between two related quantities.

Step-by-step explanation:

Often, the rate has associated units. Often, the word "per" is used to separate the quantities of the ratio, as in "miles per hour" or "dollars per gallon". In this context, "per" means "divided by."

If the units of the quantities are the same, they cancel, and the rate is a "pure number" (a number with no units). A tax rate, for example, is some number of dollars per dollar, a pure number, often expressed as a percentage.

___

Unit rates

A "unit rate" is a rate in which one of the quantities is 1 unit. Usually, that is the denominator quantity. A rate that is not a unit rate can be made to be a unit rate by carrying out the division of the numbers.

For example, 3 dollars for 2 pounds ($3/(2#)) is expressed as the unit rate $1.50 per pound.

Some years ago, grocery stores began putting unit rates on price tags so that prices could be compared more easily (at least some of the time). Sometimes the comparison is complicated by different units being used for similar products.

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Percentages

A percentage is the ratio of similar measurements, expressed with a denominator of 100. ("Cent" means "hundred" in "per cent.") The "/100" in the ratio is generally abbreviated as the symbol "%". Since the ratio is of quantities with similar units, it is a pure number.

Occasionally, you will find the idea of "percent" used to relate quantities that are measured differently. For example, a drug that has a concentration of x mg/(100 mL) may be specified as an x% solution.

The proportion of items of significantly different density may be specified either by weight or by volume. That is a mixture that is x% "by weight" may be y% "by volume" (x≠y). The choice of weight or volume will generally depend on the typical way an amount of the mixture is measured.

Answer:

Step-by-step explanation:

La tasa es un coeficiente que expresa la relación entre la cantidad y la frecuencia de un fenómeno o un grupo de números. Se utiliza para indicar la presencia de una situación que no puede ser medida en forma directa.

You mix the letters S, E, M, I, T, R, O, P, I, C, A, and L thoroughly. Without looking, you draw one letter. Find the probability that you select a vowel. Write your answer as a fraction in simplest form.

Answers

5/12 would be your simplest form
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Caleb bought groceries and paid \$1.60$1.60dollar sign, 1, point, 60 in sales tax. The sales tax rate is 2.5\%2.5%2, point, 5, percent.What was the price of Caleb's groceries, before tax?

Answers

The price before tax would be 1.60/0.025 which equals $64.

The Labor Bureau wants to estimate, at a 90% confidence level, the proportion of all households that receive welfare. A preliminary sample showed that 17.5% of households in this sample receive welfare. The sample size that would limit the margin of error to be within 0.025 of the population proportion is:_________.

Answers

Answer:

Sample size should be atleast 625

Step-by-step explanation:

Given that the  Labor Bureau wants to estimate, at a 90% confidence level, the proportion of all households that receive welfare

Sample proportion = 17.5%

Let n be the sample size

Standard error of sample proportion= \sqrt{(pq)/(n) } =\sqrt{(0.175*0.825)/(n) }

Z critical for 90% = 1.645

Margin of error = 1.645 * std error

Since margin of error<0.025 we have

1.645*\sqrt{(0.175*0.825)/(n) }<0.025\n0.625046/0.025 <√(n) \nn>625

Final answer:

The sample size that would limit the margin of error to be within 0.025 of the population proportion is approximately 185.

Explanation:

To estimate the sample size needed to limit the margin of error within 0.025, we can use the formula for sample size in proportion estimation. The formula is:

n = (Z^2 * p * (1-p)) / (E^2)

Where:

n = sample size

Z = Z-score corresponding to the desired confidence level

p = preliminary sample proportion

E = margin of error

Given that the confidence level is 90%, the Z-score for a 90% confidence level is approximately 1.645. The preliminary sample proportion is 17.5% (or 0.175) and the margin of error is 0.025.

Substituting these values into the formula:

n = (1.645^2 * 0.175 * (1 - 0.175)) / (0.025^2)

Simplifying the equation:

n = 1.645^2 * 0.175 * 0.825 / 0.025^2

n ≈ 185.16

So, the sample size that would limit the margin of error to be within 0.025 of the population proportion is approximately 185, rounded up to the nearest whole number.

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Short-term classes: Does taking a class in a short-term format (8 weeks instead of 16 weeks) increase a student’s likelihood of passing the course? For a particular course, the pass rate for the 16-week format is 59%. A team of faculty examine student data from 40 randomly selected accelerated classes and determine that the pass rate is 78%. Which of the following are the appropriate null and alternative hypotheses for this research question?a.H0: p = 0.59; Ha: p ≠ 0.59
b.H0: p = 0.59; Ha: p > 0.59
c.H0: p = 0.78; Ha: p ≠ 0.78
d.H0: p = 0.78; Ha: p > 0.59

Answers

Answer:

Option b

Step-by-step explanation:

Given that For a particular course, the pass rate for the 16-week format is 59%. A team of faculty examine student data from 40 randomly selected accelerated classes and determine that the pass rate is 78%.

we have to find whether  taking a class in a short-term format (8 weeks instead of 16 weeks) increase a student’s likelihood of passing the course

Since pass rate for 16 weeks is 59% we check whether for 8 weeks is more than this.

So correct null and alternative hypotheses for this research question

would be:

b.H0: p = 0.59; Ha: p > 0.59

Option b is right.
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