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bogdanovich [222]

1 year ago

7

Carla is shopping for Storage Bags. At the grocery store, Brand A Storage Bags are priced $4.59 for 40 count, or Brand B Storage

Bags priced $3.99 for 30 count. Which is the better buy, Brand A or Brand B?

2 answers:

vichka [17]1 year ago

8 0

Answer:

Brand A

Step-by-step explanation:

Comment

The question is one of finding out what the unit cost is.

Brand A: $4.59 has a count of 40

Brand B: $3.99 has a count of 30

Brand A

unit = 4.59 / 40 = 0.115

<em>Rounded: 0.12</em>

Brand B

unit = 3.99 / 30 = 0.133

<em>rounded: 0.13</em>

Brand A even with proper rounding is the better buy

hammer [34]1 year ago

8 0

Answer:

Brand A

Step-by-step explanation:

The cost comparison can be done a variety of ways. The most formal of solutions will calculate the cost per bag for the two brands.

<h3>Informal solution</h3>

The cost for 10 bags of Brand B is $3.99/3 = $1.33. The cost for 10 more bags (40 bags instead of 30) of Brand A is $4.59 -3.99 = $0.60. This is less than the $1.33 it would cost to buy 10 more bags of Brand B, so ...

Brand A is the better buy

<h3>Formal solution</h3>

The cost per bag is found by dividing the cost by number of bags:

Brand A: $4.59 / (40 bags) = $0.11475 per bag

Brand B: $3.99 / (30 bags) = $0.133 per bag

13¢ per bag is more than 11¢ per bag, so Brand A is the better buy.

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Let X be a Bernoulli rv with pmf as in Example 3.18. a. Compute E(X2 ). b. Show that V(X) 5 p(1 2 p). c. Compute E(X79).

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The value of E(x2) is p
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<h3>How to compute E(x2)</h3>

The distribution is given as:

p(0) = 1 - p

p(1) = p

The expected value of x2, E(x2) is calculated as:

$E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * (1- p) + 1^2 * p$

Evaluate the exponents

$E(x^2) = 0 * (1- p) + 1 * p$

Multiply

$E(x^2) = 0 +p$

Add

$E(x^2) = p$

Hence, the value of E(x2) is p

<h3>How to compute V(x)</h3>

This is calculated as:

$V(x) = E(x^2) - (E(x))^2$

Start by calculating E(x) using:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * (1- p) + 1 * p$

$E(x) = p$

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$V(x) = E(x^2) - (E(x))^2$

So, we have:

$V(x) = p - p^2$

Factor out p

$V(x) = p(1 - p)$

Hence, the value of V(x) is p(1 - p)

<h3>How to compute E(x79)</h3>

The expected value of x79, E(x79) is calculated as:

$E(x^{79}) = \sum x^{79} * P(x)$

So, we have:

$E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p$

Evaluate the exponents

$E(x^{79}) = 0 * (1- p) + 1 * p$

Multiply

$E(x^{79}) = 0 + p$

Add

$E(x^{79}) = p$

Hence, the value of E(x79) is p

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