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2 years ago

Need help on Part A!

Mathematics

2 answers:

OLga [1]2 years ago

6 0

In part A, you have been tasked with constructing an algebraic expression that represents the total cost of kitten supplies.

There are two types of kittens identified in the question: young kittens and older kittens. Let's say m is the number of young kittens and n is the number of older kittens.

The young kittens' food is $3.99 per bottle. The older kittens' food is $1.50 per can.

$3.99m + 1.50n$

Based on the number of each type of kitten, the cost will vary.

The second part of the question asks you to identify specific parts of the equation. The coefficients are 3.99 and 1.50. There are no constants in the expression. The variables are m and n.

3241004551 [841]2 years ago

4 0

Let k represent the cost of supplies, b the number of bottles, and c the number of cans. You know that the total cost is found by adding the number of bottles multiplied by the price of each to the number of cans multiplied by their unit price. (This is the computation performed anytime you purchase something.)

$\boxed{k}=\underline{3.99}\boxed{b}+\underline{1.50}\boxed{c}$

There are no constants in this equation.

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A sample of 5 buttons is randomly selected and the following diameters are measured in inches. Give a point estimate for the pop

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$s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

But we need to calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_I}{n}$

And replacing we got:

$\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072$

And for the sample variance we have:

$s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003$

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance $\sigma^2$

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Step-by-step explanation:

For this case we have the following data:

1.04,1.00,1.13,1.08,1.11

And in order to estimate the population variance we can use the sample variance formula:

$s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

But we need to calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_I}{n}$

And replacing we got:

$\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072$

And for the sample variance we have:

$s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003$

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance $\sigma^2$

$E(s^2) = \sigma^2$

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