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1 year ago

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a building block in the shape of a cube was found in a toy box . it has side lengths of 1.1 inches what is the volume of the bui

lding block cubic inches? your answer may be exact or rounded to the nearest tenth.

2 answers:

Travka [436]1 year ago

7 0

volume= a^3

= 1.1^3

= 1.331 in^3

= 1.33 in^3

Firdavs [7]1 year ago

5 0

<h3>given:</h3>

side= 1.1 inches

<h3>to find:</h3>

the volume of the given cube rounded to the nearest tenth.

<h3>solution:</h3>

$v = {a}^{3}$

$v = {1.1}^{3}$

$v = 1.331 \: {in}^{3}$

$v = 1 .33 \: {in}^{3}$

<u>therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>volume</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>cube</u><u> </u><u>is</u><u> </u><u>1.33</u><u> </u><u>cubic</u><u> </u><u>inches</u><u>.</u>

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Kay [80]

Answer:

<h2>c. x = -4 or x = 9</h2>

Step-by-step explanation:

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}$

Domain:

$x\neq0\ \wedge\ x^2-9\neq0\ \wedge\ x-3\neq0\\\\x\neq0\ \wedge\ x\neq\pm3$

solution:

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multiply both sides by (x - 3) ≠ 0

$\dfrac{4(x-3)}{x}+\dfrac{4(x-3)}{(x-3)(x+3)}=\dfrac{3(x-3)}{x-3}$

cancel (x - 3)

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subtract $\frac{4(x-3)}{x}$ from both sides

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subtract 4x from both sides

$0=-x^2+12x-3x+36-4x$

combine like terms

$0=-x^2+(12x-3x-4x)+36\\\\0=-x^2+5x+36$

change the signs

$x^2-5x-36=0\\\\x^2-9x+4x-36=0\\\\x(x-9)+4(x-9)=0\\\\(x-9)(x+4)=0$

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2 years ago

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

Helen [10]

Answer:

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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:

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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:

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

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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3 years ago

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son4ous [18]

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5 0

1 year ago

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