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3 years ago

7

Shaz is cleaning under the kitchen sink. He found five 2-liter bottles of glass cleaner. None of the bottles is full. The bottle

s contain the following amounts.
Bottle 1 – 375 mL Bottle 2 – 150 mL Bottle 3 – 190 mL
Bottle 4 – 780 mL Bottle 5 – 630 mL
How many liters of glass cleaner do the five bottles contain in total?
If Shaz combines all of the glass cleaners into as few bottles as possible, how many bottles will he use?

2 answers:

Stels [109]3 years ago

6 0

Sbshehe fjshebd Shane

elixir [45]3 years ago

6 0

Step-by-step explanation:

375 + 150 + 190 + 780 + 630 = 2125 ml

For first question

2125 ml ÷ 10^3 = 2.125 liter

And by criss cross method 2 litter - > 1 bottle

2.125 liter - > x ( x is unknown)

X= 2.125 / 2 = 1.0625 bottles

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$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

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From the box, we have:

$Least = 1$

So:

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The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

<u>Lower half:</u>

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

<u>Upper half</u>

<u></u> $\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$ <u></u>

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

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We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

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$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

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$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

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$7th = 18$

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$Median = \frac{1}{2}(7th + 8th)$

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$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

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$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

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