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NikAS [45]
3 years ago
8

A certain brand of coffee comes in two sizes. A 10.5-ounce package costs $2.99. A 29.3-ounce package costs $7.98. Find the unit

price for each size. Then state which size is the better buy based on the unit price. Round your answers to the nearest cent.
Mathematics
1 answer:
klio [65]3 years ago
5 0

Answer:

The unit price are:

$0.285 for the 10.5 ounce package

$0.272 for the 29.3 ounce package.

The size that is the better buy based on the unit price is the 29.3-ounce package costs $7.98 because it's unit's price($0.272) is lesser than that of the 10.5 ounce package.

Step-by-step explanation:

From the question:

A 10.5-ounce package costs $2.99.

The unit price = 1 ounce.

10.5 ounce = $2.99

1 ounce = x

10.5 ounce × x = 1 ounce × $2.99

x = 1 ounce × $2.99/10.5 ounce

x = $0.2847619048

The unit price = $0.2847619048

Approximately = $0.285

A 29.3-ounce package costs $7.98.

The unit price = 1 ounce.

29.3 ounce = $7.98

1 ounce = x

29.3 ounce × x = 1 ounce × $7.98

x = 1 ounce × $7.98/29.3 ounce

x = $0.2723549488

The unit price = $0.2723549488

Approximately = $0.272

The size that is the better buy based on the unit price is the 29.3-ounce package costs $7.98 because it's unit's price($0.272) is lesser than that of the 10.5 ounce package.

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Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1
AysviL [449]

Answer:

a) The system has a unique solution for k\neq 6 and any value of h, and we say the system is consisted

b) The system has infinite solutions for k=6 and h=8

c) The system has no solution for k=6 and h\neq 8

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms (h), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]

The we can use the transformation r_0\rightarrow r_0 -2r_1, obtaining:

\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right].

Now we can start the analysis:

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(k-6)x_2=h-8\\x_2=h-8/(k-6)

from where we can see that only in the case of k=6 the value of x_2 can not be determined.

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if k=6 and h\neq 8 the system has no solution. Again by substituting in the equation resulting from the last row:

(k-6)x_2=h-8\\0=h-8 which is false for all values of h\neq 8 and since we have something that is not possible (0\neq h-8,\ \forall \ h\neq 8) the system has no solution

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Step-by-step explanation:

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