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Darya [45]
3 years ago
13

A bulk store sells 200 diapers from Brand (a) for $38.75 and it says 170 diapers from Brand (b) for $35.50. If both brands have

the same quality which is the better deal ?
Mathematics
1 answer:
Mazyrski [523]3 years ago
5 0

Answer:

Brand A costs less per diaper

Step-by-step explanation:

To determine if it is the better deal, we need the cost per diaper.  To find the Cost per diaper, take the total cost and divide by the number of diapers

Brand A

Cost per diaper = $38.75/ 200 = $.19375  per diaper

Brand B

Cost per diaper = $35.50/170 =.208823529 per diaper


Brand A costs less per diaper

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kondor19780726 [428]

Answer:

30 and 36

Step-by-step explanation:

six times two equals 12

6 * 3 = 18

and six times four equals 24.

to get the next two multiples, You need to multiply 6 by 5, and then 6

6 * 5 = 30 and 6 * 6 = 36

3 0
3 years ago
The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t
shtirl [24]

Answer:

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

Step-by-step explanation:

As the given Augmented matrix is

\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 1 :

r_{1}↔r_{1} - r_{2}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 2 :

r_{3}↔r_{3} - 8r_{1}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]

Step 3 :

r_{2}↔\frac{r_{2}}{7}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]

Step 4 :

r_{1}↔r_{1} + 14r_{2} , r_{3}↔r_{3} - 124r_{2}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]

Step 5 :

r_{3}↔\frac{r_{3}. 7}{254}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]

Step 6 :

r_{1}↔r_{1} - 4r_{3} , r_{2}↔r_{2} + \frac{1}{7} r_{3}

\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]

∴ we get

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

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Answer: A.

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