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lara31 [8.8K]
3 years ago
7

The weekly ad for a local grocery store advertises a 5 pound bag of organic apples for $13.95. Round each rate to the nearest hu

ndredth if needed. Find both unit rates. Which unit rate is typically used? Explain.
Mathematics
2 answers:
BigorU [14]3 years ago
7 0

Given values are :

5 pound of organic apple costs $13.95

So, 1 pound of organic apple will cost = \frac{13.95}{5}= 2.79

So, per unit rate of apples will be $2.79 per pound.

We will also find the another unit rate of per dollar.

So, using $13.95 we can purchase = 5 pounds of apples

Hence, with $1 we can purchase = \frac{5}{13.95}= 0.36 pounds of apples.

So, per unit money (dollars) the amount of apples will be 0.36 pounds.

Here, we can say that the 1st part;  the per unit rate of apples that depicts how many apples can be bought in $2.79 is the unit rate used.

beks73 [17]3 years ago
3 0

Answer:

$2.79 per pound of apples.

0.36 pounds per dollar.

The unit rate representing cost of per pound apples is typically used.

Step-by-step explanation:

We have been given that the weekly ad for a local grocery store advertises a 5 pound bag of organic apples for $13.95.

Since we know when rates are expressed as a quantity of 1, such as 100 meter per second or 5 miles per hour, they are called unit rates. We need to have 1 in our denominator to express two quantities as unit rate.

Let us find the unit rate in terms of cost of per pound bag of apples.

\text{Unit rate as the cost of organic apples per pound}=\frac{\$13.95}{5 \text{ pounds of organic apples}}

\text{Unit rate as the cost of organic apples per pound}=\frac{\$2.79} {\text{ pound of organic apples}}

Therefore, our unit rate will be $2.79 per pound of apples.

Let us find unit rate of pounds of apples per dollar.

\text{Unit rate as the pounds of organic apples per dollar}=\frac{5\text{ pounds of organic apples}}{\$13.95}

\text{Unit rate as the pounds of organic apples per dollar}=\frac{0.3584 \text{ pounds of organic apples}}{\$}

\text{Unit rate as the pounds of organic apples per dollar}\approx \frac{0.36 \text{ pounds of organic apples}}{\$}

Therefore, our another unit rate will be 0.36 pounds per dollar.

Since, we purchase apples or other items according to their price per piece or their price per pound in our daily life, therefore, the unit rate representing cost of per pound apples is typically used.

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Step-by-step explanation:

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Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival ti
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Answer:

D. 91%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5

Simplifying:

f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

P(A) = \int_{5}^{15} = \frac{375}{t^4}

Integral of \frac{1}{t^4} = t^{-4} is \frac{t^{-3}}{-3} = -\frac{1}{3t^3}

Then

\int \frac{375}{t^4} dt = -\frac{125}{t^3}

Applying the limits, by the Fundamental Theorem of Calculus:

At t = 15, f(15) = -\frac{125}{15^3} = -\frac{1}{27}

At t = 5, f(5) = -\frac{125}{5^3} = -1

Then

P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

P(B) = \int_{5}^{10} = \frac{375}{t^4}

We already have the integral, so just apply the limits:

At t = 10, f(10) = -\frac{125}{10^3} = -\frac{1}{8}

At t = 5, f(5) = -\frac{125}{5^3} = -1

Then

P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

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