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yan [13]
2 years ago
10

(PLEASE HELP) To feed 7 horses for 4 days, Mark needs 420 kg of grass. How many kilograms of grass does Mark need to feed 9 hors

es for 6 days?
Mathematics
2 answers:
DerKrebs [107]2 years ago
7 0

Answer:

810kg is the answer

Step-by-step explanation:

Hope this helps:)

Vlada [557]2 years ago
6 0
420 divided by 4 is 105 so each day 7 horses eat 105 kg of grass,
105 divided by 7 = 15 so each horse intakes 15 kg of grass per day

9x15=135
135x6 =810
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Answer:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

M = \frac{1}{2}(Lower + Upper)

Where

Lower \to Lower class interval

Upper \to Upper class interval

So, we have:

Class 63-65:

M = \frac{1}{2}(63 + 65) = 64

Class 66 - 68:

M = \frac{1}{2}(66 + 68) = 67

When the computation is completed, the frequency distribution will be:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

\bar x = 70.2903

\sigma = 3.5795

<em>See attachment for result of 1-VarStats</em>

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Answer:

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Hope you have a nice day and understood this!

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