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Artemon [7]
3 years ago
14

A sample of small bottles and their contents has the following weights (in grams): 4, 2, 5, 4, 5, 2, and 6. What is the sample v

ariance of bottle weight
Mathematics
1 answer:
Inessa05 [86]3 years ago
3 0

Answer:

What is the sample variance of bottle weight 2.33

Step-by-step explanation:

First find the mean. The mean of the bottle weight is obtained by taking the ratio of the sum of ages and total number of ages.

Mean = (4+2+5+4+5+2+6) / 7

          = 28 / 7

          = 4

Sample Variance = \frac{\left[\begin{array}{ccc}(4-4)^{2} +(2-4)^{2} +\\(5-4)^{2} + (4-4)^{2} +\\(5-4)^{2} + (2-4)^{2} +\\(6-4)^{2}\end{array}\right]}{7-1}

                             =\frac{0+4+1+0+1+4+4}{6}

                             = 2.33

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Answer:

-3119

Step-by-step explanation:

(-5)^2 - 2 x (-9) + 6

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-3125 - 2 x -9 + 6

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3 years ago
IVedle Question Help Miranda's financial aid stipulates that her tuition not exceed $1500. If her college charges a $45 registra
forsale [732]

Answer:

Therefore, Miranda can register for at most 3 courses.

Explanation:

If x is the number of courses in which Miranda register, the total cost for her tuition will be:

45 + 450x

Because there is a fixed cost of $45 and a cost per course of $450.

Then, this cost should be less than or equal to $1500, so we can write the following inequality:

45 + 450x ≤ 1500

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7 0
1 year ago
Belinda is making doughnuts. If the recipe calls for 2 cups of sugar, and each cup of sugar weighs 2/3 pounds, how many pounds o
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Step-by-step explanation:

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3 years ago
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Answer:

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3 years ago
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The length and width of a rectangle are measured as 30cm and 24cm, respectively, with an error in measured of
larisa [96]

Answer:

The maximum error in the calculated area of rectangle is 5.4 cm².

Step-by-step explanation:

Given : The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measured of  at most 0.1 cm in each.

To find : Use differentials to estimate the maximum error in the calculated area of rectangle ?

Solution :

The area of the rectangle is A=l\times w

The derivative of the area is equal to the partial derivative of area w.r.t. length times the change in length plus the partial derivative of area w.r.t. width times the change in width.

i.e. dA=\frac{\partial A}{\partial L}\Delta L+\frac{\partial A}{\partial W}\Delta W

Here, \frac{\partial A}{\partial L}=30,\ \frac{\partial A}{\partial W}=24 ,\ \Delta L=0.1,\ \Delta W=0.1

Substitute the values,

dA=(30)(0.1)+(24)(0.1)

dA=3+2.4

dA=5.4

Therefore, the maximum error in the calculated area of rectangle is 5.4 cm².

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