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oksian1 [2.3K]
3 years ago
13

How many 3/4 cup servings are in 12 cups of cherries?

Mathematics
1 answer:
AveGali [126]3 years ago
8 0

Answer:

16 serving.

Step-by-step explanation:

Given: 3/4 cup servings.

Now, computing the number of serving in 12 cups of cherries.

Using unitary method to solve.

∴ Number of serving= total\ cups \times number\ of\ serving\ per\ cup

Number of serving in one cup is \frac{4}{3}

Number of serving in 12 cups = 12\times \frac{4}{3} = 16 \ serving

Number of serving in 12 cups= 16\ serving

∴ There are 16 serving is possible in 12 cups of cherries.

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5 0
3 years ago
Show that there do not exist scalar c1 c2 and c3 c1(-2,9,6)+c2(-3,2,1)+c3(1,7,5)= (0,5,4)​
fredd [130]

This is the same as showing the following system of equations doesn't have a solution:

\begin{cases}-2c_1-3c_2+c_3 = 0 \\ 9c_1 + 2c_2 + 7c_3 = 5 \\ 6c_1 + c_2 + 5c_3 = 4\end{cases}

or in matrix form,

\begin{bmatrix}-2&9&6\\-3&2&1\\1&7&5\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix} = \begin{bmatrix}0\\5\\4\end{bmatrix}

The quickest way to check if there is a solution is to check whether the coefficient matrix is invertible. If its determinant is 0, then it is not invertible.

And the quickest way to show that the determinant is 0 is by observing that the third row is a linear combination of the first two rows:

(-2, 9, 6) - (-3, 2, 1) = (-2 + 3, 9 - 2, 6 - 1) = (1, 7, 5)

So there are indeed no such scalars <em>c₁</em>, <em>c₂</em>, and <em>c₃</em>.

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3 years ago
Evaluate the expression of 26^3+5 when b=3
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Answer:

i dont knoy sorry i am in hurry

5 0
2 years ago
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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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Answer:

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

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:

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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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Answer:

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