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3 years ago

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Problem PageQuestion Tony estimated the height of his office building to be 45ft. The actual height of his office building is 41

ft. Find the absolute error and the percent error of Tony's estimate. If necessary, round your answers to the nearest tenth.

1 answer:

Leona [35]3 years ago

5 0

Answer:

The absolute error is <u>4 ft</u> and percent error of Tony's estimate is <u>9.8%</u>.

Step-by-step explanation:

Given:

Tony estimated the height of his office building to be 45ft. The actual height of his office building is 41ft.

Now, to find the absolute error and the percent error of Tony's estimate.

The height of his office building = 45ft.

The actual height of his office building = 41 ft.

So, to get the absolute error:

<em>The height of his office building - the actual height of his office building</em>

$45\ ft-41\ ft\\=4\ ft.$

The absolute error = 4 ft.

Now, to get the percent error of Tony's estimate:

$\frac{4}{41} \times 100$

$=0.0975\times 100$

$=9.75\%.$

<em>Percent error of Tony's estimate rounding to nearest tenth = 9.8%.</em>

Therefore, the absolute error is 4 ft and percent error of Tony's estimate is 9.8%.

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Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

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Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

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2 years ago

20. A ball is dropped from a height of a little over 5 feet, and the height is measured at small intervals. The table below show

nekit [7.7K]

Step 1:

a) Write the quadratic model

a = -15.64

b = -1.24

c = 5.23

$P(t)=-15.64t^2\text{ - 1.24t + 5.2}3$

b) t = 0.30

$\begin{gathered} P(t)=-15.64t^2\text{ - 1.24t + 5.2}3 \\ =\text{ -15.64 }\times0.3^2\text{ - 1.24}\times\text{ 0.3 + 5.2}3 \\ =\text{ -1.4076 - 0.372 + }5.23 \\ =\text{ 3.45} \end{gathered}$

c) t = 0.52 seconds

$\begin{gathered} p(t)=-15.64t^2\text{ - 1.23t + 5.23} \\ =\text{ -15.64}\times0.52^2\text{ - 1.23}\times0.52\text{ + 5.23} \\ =\text{ -4.23 - 0.64 + 5.23} \\ =\text{ 0.36} \end{gathered}$

d) 0.30 is more likely to be relaible.

e)

$\begin{gathered} p(t)\text{ = 1} \\ p(t)=-15.64t^2\text{ - 1.23t + 5.23} \\ 1=-15.64t^2\text{ - 1.23t + 5.23} \\ -15.64t^2\text{ - 1.23t + 4.23 = 0} \\ t\text{ = 0.48222} \\ \text{t = 0.48 seconds} \end{gathered}$

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1 year ago