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3 years ago

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Vince’s family runs three lemonade stands during local events. Each stand serves lemonade in small, medium, and large glasses. T

he table gives the number of glasses of each size sold in each stand. Small Medium Large Stand 1 small 12 medium 11 large 6 Stand 2 small 10 medium 6 large 1 Stand 3 small 0 medium 3 large 9 The total collected from each stand was: stand 1, $87; stand 2, $43; and stand 3, $54. What is the price of a medium glass of lemonade? A. $2 B. $3 C. $4.50 D. $5

2 answers:

nalin [4]3 years ago

5 0

Answer:

B. $3

Step-by-step explanation:

Let s, m, l represent the prices of small, medium, and large lemonades, respectively. Three equations can be written based on the given revenue:

12s +11m +6l = 87
10s +6m +1l = 43
0s +3m +9l = 54

We can use the last equation to write an expression for l:

9l = 54 -3m . . . subtract 3m

l = 6 -m/3 . . . . . divide by 9

Substituting this into the first two equations gives ...

12s +11m +6(6 -m/3) = 87 . . . . . sub for l in first equation

12s +9m = 51 . . . . subtract 36, collect terms

Dividing by 3 gives

4s +3m = 17

__

10s +6m +(6 -m/3) = 43

10s + (5 1/3)m = 36 . . . . subtract 6, collect terms

30s +16m = 108 . . . . . . multiply by 3

Dividing by 2 gives

15s + 8m = 54

__

We can multiply the first of these by 15 and the second by -4, then add the results to eliminate the s variable.

15(4s +3m) -4(15s +8m) = 15(17) -4(54)

60s +45m -60s -32m = 255 -216 . . . . . eliminate parentheses

13m = 39 . . . . . collect terms

m = 3 . . . . . . . . divide by 13

The price of a medium glass of lemonade is $3.

_____

Most graphing calculators and available on-line calculators can make solving three equations in three unknowns trivially simple.

MAVERICK [17]3 years ago

3 0

Answer:

3$

Step-by-step explanation:

I did the Plato test

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A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the numb

Maksim231197 [3]

Answer:

a) $P(X = 1) = 0.38742$

b) $P(X = 3) = 0.05740$

c) $P(X = 9) = 0.00000$

d) $P(X \geq 5) = 0.00163$

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem

There are 10 containers, so $n = 10$ .

A food-packaging apparatus underfills 10% of the containers, so $p = 0.1$ .

a) This is P(X = 1)

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

b) This is P(X = 3)

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

c) This is P(X = 9)

$P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000$

d) This is $P(X \geq 5)$ .

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

$P(X < 5) + P(X \geq 5) = 1$

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868$

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

$P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937$

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

$P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742$

So

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837$

Finally

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163$

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