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3 years ago

7

Chester’s bakery a box of 12 donuts sells for 8.50 what is the unit price of donuts in the box rounded to the nearest cent

1 answer:

Firdavs [7]3 years ago

3 0

Unit price = price for one donut
if 12 donuts cost $8.50, then one donut would be:
$8.50/12 = 0.7083, or about $0.71

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A deli charges the same price, $5.40, for all of its different sandwiches. If a company ordered n sandwiches for a luncheon, whi

zaharov [31]

P=5.40n It is the correct answer

6 0

3 years ago

Weary of the low turnout in student elections, a college administration decides to choose an SRS of three students to form an ad

-Dominant- [34]

Answer:

$P(ABC) = 0.110592$

$P(ABC^c) = 0.119808$

$P(AB^cC) = 0.119808$

$P(A^cBC) = 0.119808$

$P(AB^cC^c) = 0.129792$

$P(A^cBC^c) = 0.129792$

$P(A^cB^cC) = 0.129792$

$P(A^cB^cC^c) = 0.140608$

Step-by-step explanation:

Given

$P(A) = P(B) = P(C) = 48\%$

Convert the probability to decimal

$P(A) = P(B) = P(C) = 0.48$

Solving (a): P(ABC)

This is calculated as:

$P(ABC) = P(A) * P(B) * P(C)$

This gives:

$P(ABC) = 0.48*0.48*0.48$

$P(ABC) = 0.110592$

Solving (b): $P(ABC^c)$

This is calculated as:

$P(ABC^c) = P(A) * P(B) * P(C^c)$

In probability:

$P(C^c) = 1 - P(C)$

So, we have:

$P(ABC^c) = P(A) * P(B) * (1 - P(C))$

$P(ABC^c) = 0.48 * 0.48 * (1 - 0.48)$

$P(ABC^c) = 0.48 * 0.48 * 0.52$

$P(ABC^c) = 0.119808$

Solving (c): $P(AB^cC)$

This is calculated as:

$P(AB^cC) = P(A) * P(B^c) * P(C)$

$P(AB^cC) = P(A) * [1 - P(B)] * P(C)$

$P(AB^cC) = 0.48 * (1 - 0.48)* 0.48$

$P(AB^cC) = 0.48 * 0.52* 0.48$

$P(AB^cC) = 0.119808$

Solving (d): $P(A^cBC)$

This is calculated as:

$P(A^cBC) = P(A^c) * P(B) * P(C)$

$P(A^cBC) = [1-P(A)] *P(B) * P(C)$

$P(A^cBC) = (1 - 0.48)* 0.48 * 0.48$

$P(A^cBC) = 0.52* 0.48 * 0.48$

$P(A^cBC) = 0.119808$

Solving (e): $P(AB^cC^c)$

This is calculated as:

$P(AB^cC^c) = P(A) * P(B^c) * P(C^c)$

$P(AB^cC^c) = P(A) * [1-P(B)] * [1-P(C)]$

$P(AB^cC^c) = 0.48 * [1-0.48] * [1-0.48]$

$P(AB^cC^c) = 0.48 * 0.52*0.52$

$P(AB^cC^c) = 0.129792$

Solving (f): $P(A^cBC^c)$

This is calculated as:

$P(A^cBC^c) = P(A^c) * P(B) * P(C^c)$

$P(A^cBC^c) = [1-P(A)] * P(B) * [1-P(C)]$

$P(A^cBC^c) = [1-0.48] * 0.48 * [1-0.48]$

$P(A^cBC^c) = 0.52 * 0.48 * 0.52$

$P(A^cBC^c) = 0.129792$

Solving (g): $P(A^cB^cC)$

This is calculated as:

$P(A^cB^cC) = P(A^c) * P(B^c) * P(C)$

$P(A^cB^cC) = [1-P(A)] * [1-P(B)] * P(C)$

$P(A^cB^cC) = [1-0.48] * [1-0.48] * 0.48$

$P(A^cB^cC) = 0.52 * 0.52 * 0.48$

$P(A^cB^cC) = 0.129792$

Solving (h): $P(A^cB^cC^c)$

This is calculated as:

$P(A^cB^cC^c) = P(A^c) * P(B^c) * P(C^c)$

$P(A^cB^cC^c) = [1-P(A)] * [1-P(B)] * [1-P(C)]$

$P(A^cB^cC^c) = [1-0.48] * [1-0.48] * [1-0.48]$

$P(A^cB^cC^c) = 0.52*0.52*0.52$

$P(A^cB^cC^c) = 0.140608$

5 0

3 years ago

alexira [117]

Answer: when x = 7, y = 16

Step-by-step explanation: Here, we know that y varies inversely as x.

When we have two sets of inversely related coordinates, x₁ and y₁,

and x₂ and y₂, we can use the product rule, shown below,

to find the missing value.

<h2>x₁ y₁ = x₂ y₂</h2><h2 />

Here, we know that y = 14 when x = 8 so

one set of coordinates will be 8 and 14.

We want to know the value of y when x = 7.

So our other coordinates will be 7 and y.

So we have (8)(14) = (7)(y).

Simplifying, (8)(14) is equal to 112 and (7)(y) is equal to 7y.

So we have 112 = 7y.

Now we simply solve for y by dividing both sides of the equation by 7.

The 7's on the right side cancel out and

on the left, 112 divided by 7 is equal to 16.

So we have 16 = y.

So when x is equal to 7, y is equal to 16.

7 0

3 years ago

Leon read 5/7 of the pages in his book.He has 28 pages left to read.How many pages has he read already?

Fittoniya [83]

You would first divide 28 by 7.
28 / 7 = 4.
Now multiply 4 by 5.
4 X 5 = 20.

Therefore, Leon read 20 pages.

4 0

3 years ago

119 divided by 17? with remainder?

Pachacha [2.7K]

Answer:

A number, when divided by 119, leaves a remainder of 19. If it is divided by 17, it will leave a remainder of 2

Step-by-step explanation:

5 0

4 years ago