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3 years ago

Give me example of a unit rate used in real life

Mathematics

1 answer:

lana66690 [7]3 years ago

3 0

Before you begin this lesson, please print the accompanying document, Unit Rates in Everyday Life].

Have you ever been at the grocery store and stood, staring, at two different sizes of the same item wondering which one is the better deal? If so, you are not alone. A UNIT RATE could help you out when this happens and make your purchasing decision an easy one.

In this lesson, you will learn what UNIT RATES are and how to apply them in everyday comparison situations. Click the links below and complete the appropriate sections of the Unit Rates handout.

[Note: The links below were created using the Livescribe Pulse Smartpen. If you have never watched Livescribe media before, take a few minutes to watch this very brief Livescribe orientation]

What is a UNIT RATE – definitionView some examples of Unit RatesSee a process to compute Unit Rates

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Don ate 160 grams of carbohydrates, 100 grams of fat, and 70 grams of protein today. What is his total calorie consumption for t

Artemon [7]

Answer:

His total calorie consumption for the day is:

Option C. 1.820 kcal

Step-by-step explanation:

Protein and carbohydrates both contain 4 calories per gram, while fat provides 9 calories per gram, then:

1.) 160 grams of carbohydrates contain:

160 grams * (4 cal/gram)=640 cal

2.) 100 grams of fat contain:

100 grams * (9 cal/gram)=900 cal

3.) 70 grams of protein contain:

70 grams * (4 cal/gram)=280 cal

Then, his total calorie consumption for the day is:

T=640 cal + 900 cal +280 cal

T=1,820 cal

T=1,820 cal (1 kcal / 1,000 cal)

T=1.820 kcal

6 0

3 years ago

Simplify the expression 8x-2x+12x-3

aev [14]

Answer:

18x - 3

Step-by-step explanation:

Add all the X values

8x + -2x + 12x = 18x

Put the final numbers on the side.

-3

18x-3

7 0

2 years ago

Thee population of a town, P(t), is 2000 in the year 2010. Every year, it increases by 1.5%. Write an equation for the populatio

steposvetlana [31]

Given:

In 2010, the population of a town = 2000

Every year, it increase by 1.5%

To find the equation P(t) which represents the population of this town t years from 2010.

Formula

If a be the original population of a town and it increases by b% every year, after t year the population will be

$a (1+\frac{b}{100} )^{t}$

Now,

Taking, a =2000, b = 1.5 we get,

$P(t) = 2000(1+\frac{1.5}{100} )^{t}$

or, $P(t) = 2000(1+0.015)^{t}$

Hence,

The population of this town t years from 2010 P(t) = $2000(1+0.015)^{t}$ , Option D.

7 0

3 years ago

Brenda has money invested in Esti Transport. She owns two par value $1,000 bonds issued by Esti Transport, which currently sells

Debora [2.8K]

Answer:

B is the the answer but I'm not sure

7 0

3 years ago

An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

$P(X = x) = \frac{y^xe^{-y}}{x!}$

Where y represents λ (average)

y = 2.3

Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment

Using probability notations;

$P(X\leq 1) = P(X=0) + P(X =1)$

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

$P(X = 0) = \frac{1 * e^{-2.3}}{1}$

$P(X = 0) = e^{-2.3}$

$P(X = 0) = 0.10025884372$

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

$P(X = 1) =2.3 * e^{-2.3}$

$P(X = 1) = 2.3 * 0.10025884372$

$P(X = 1) = 0.23059534055$

$P(X\leq 1) = P(X=0) + P(X =1)$

$P(X\leq 1) = 0.10025884372 + 0.23059534055$

$P(X\leq 1) = 0.33085418427$

$P(X\leq 1) = 0.331$

Hence, the required probability is 0.331

6 0

2 years ago