Cabiolab
2021-01-02
Answered

45% of s is 18

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broliY

Answered 2021-01-03
Author has **97** answers

To find the percent of a number, you multiply the percent in decimal form and the number.Therefore 45% of s is 0.45s.

If 45% of s is 18, then

To solve for s, you then need to multiply both sides of the equation by

Therefore,

asked 2021-05-14

Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.

$\begin{array}{|ccccccc|}\hline 11.8& 7.7& 6.5& 6.8& 9.7& 6.8& 7.3\\ 7.9& 9.7& 8.7& 8.1& 8.5& 6.3& 7.0\\ 7.3& 7.4& 5.3& 9.0& 8.1& 11.3& 6.3\\ 7.2& 7.7& 7.8& 11.6& 10.7& 7.0\\ \hline\end{array}$

a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion.$[Hint.\text{}?{x}_{j}=219.5.]$ (Round your answer to three decimal places.)

MPa

State which estimator you used.

$x$

$p?$

$\frac{s}{x}$

$s$

$\stackrel{~}{\chi}$

b) Calculate a point estimate of the strength value that separates the weakest$50\mathrm{\%}$ of all such beams from the strongest $50\mathrm{\%}$ .

MPa

State which estimator you used.

$s$

$x$

$p?$

$\stackrel{~}{\chi}$

$\frac{s}{x}$

c) Calculate a point estimate of the population standard deviation ?.$[Hint:\text{}?{x}_{i}2=1859.53.]$ (Round your answer to three decimal places.)

MPa

Interpret this point estimate.

This estimate describes the linearity of the data.

This estimate describes the bias of the data.

This estimate describes the spread of the data.

This estimate describes the center of the data.

Which estimator did you use?

$\stackrel{~}{\chi}$

$x$

$s$

$\frac{s}{x}$

$p?$

d) Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa. [Hint: Think of an observation as a "success" if it exceeds 10.] (Round your answer to three decimal places.)

e) Calculate a point estimate of the population coefficient of variation$\frac{?}{?}$ . (Round your answer to four decimal places.)

State which estimator you used.

$p?$

$\stackrel{~}{\chi}$

$s$

$\frac{s}{x}$

$x$

a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion.

MPa

State which estimator you used.

b) Calculate a point estimate of the strength value that separates the weakest

MPa

State which estimator you used.

c) Calculate a point estimate of the population standard deviation ?.

MPa

Interpret this point estimate.

This estimate describes the linearity of the data.

This estimate describes the bias of the data.

This estimate describes the spread of the data.

This estimate describes the center of the data.

Which estimator did you use?

d) Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa. [Hint: Think of an observation as a "success" if it exceeds 10.] (Round your answer to three decimal places.)

e) Calculate a point estimate of the population coefficient of variation

State which estimator you used.

asked 2022-06-13

Help with simplification of a rational expression (with fractional powers)

Can you please help me see what I don't see yet. Here's a problem from a high school textbook (ISBN 978-5-488-02046-7 p.9, #1.029):

$\frac{({a}^{1/m}-{a}^{1/n}{)}^{2}\cdot 4{a}^{(m+n)/mn}}{({a}^{2/m}-{a}^{2/n})(\sqrt[m]{{a}^{m+1}}+\sqrt[n]{{a}^{n+1}})}$

Here's my try at it:

$\frac{({a}^{1/m}-{a}^{1/n})({a}^{1/m}-{a}^{1/n})\cdot 4{a}^{(1/m)+(1/n)}}{({a}^{1/m}-{a}^{1/n})({a}^{1/m}+{a}^{1/n})\cdot a({a}^{1/m}+{a}^{1/n})}$

...which is

$\frac{({a}^{1/m}-{a}^{1/n})\cdot 4{a}^{(1/m)+(1/n)}}{a({a}^{1/m}+{a}^{1/n}{)}^{2}}$

Wolfram Alpha's simplify stops here, too. I don't see where to go from here. The final form, according to the book, is this:

$\frac{1}{a({a}^{1/m}-{a}^{1/n})}$

How did they do it?

Can you please help me see what I don't see yet. Here's a problem from a high school textbook (ISBN 978-5-488-02046-7 p.9, #1.029):

$\frac{({a}^{1/m}-{a}^{1/n}{)}^{2}\cdot 4{a}^{(m+n)/mn}}{({a}^{2/m}-{a}^{2/n})(\sqrt[m]{{a}^{m+1}}+\sqrt[n]{{a}^{n+1}})}$

Here's my try at it:

$\frac{({a}^{1/m}-{a}^{1/n})({a}^{1/m}-{a}^{1/n})\cdot 4{a}^{(1/m)+(1/n)}}{({a}^{1/m}-{a}^{1/n})({a}^{1/m}+{a}^{1/n})\cdot a({a}^{1/m}+{a}^{1/n})}$

...which is

$\frac{({a}^{1/m}-{a}^{1/n})\cdot 4{a}^{(1/m)+(1/n)}}{a({a}^{1/m}+{a}^{1/n}{)}^{2}}$

Wolfram Alpha's simplify stops here, too. I don't see where to go from here. The final form, according to the book, is this:

$\frac{1}{a({a}^{1/m}-{a}^{1/n})}$

How did they do it?

asked 2021-07-31

A number that has exactly 2 factors (1 and itself) a called a _______ number

1) factors

2) prime

3) composite

4) multiplication

1) factors

2) prime

3) composite

4) multiplication

asked 2021-09-26

A population of bacteria is initially 6000. After three hours the population is 3000.

If this rate of decay continues, find the exponential function that represents the size of the bacteria population after t hours. Write your answer in the form$f\left(t\right)=a\left(b\right)t$ .

Use either fractions, or decimals rounded to 4 places

If this rate of decay continues, find the exponential function that represents the size of the bacteria population after t hours. Write your answer in the form

Use either fractions, or decimals rounded to 4 places

asked 2022-05-21

I have an electronic weighing-machine, which I believe to be internally very accurate. It will weigh up to 100 kg, but not activate below 10 kg. The digital display reports to one decimal place. The problem is that I don't know whether the reading is rounded (with worst error $\pm 50$50 g) or truncated (with worst error −100 g and expected bias −50 g). I have a great quantity of books and papers that can be stacked on the machine to make any weight within its limits, but nothing of accurately known weight.

I guess that any solution must be statistical. A good solution would minimize the number of weighings, given a tolerance probability of a false indication. (Assume a 50/50 prior distribution for rounding/truncation; for illustration, a targeted probability could be 0.1%.)

I guess that any solution must be statistical. A good solution would minimize the number of weighings, given a tolerance probability of a false indication. (Assume a 50/50 prior distribution for rounding/truncation; for illustration, a targeted probability could be 0.1%.)

asked 2022-02-04

How do you find the Least common multiple of

asked 2022-07-19

If $2f(x)+3f(\frac{1}{x})=\frac{4{x}^{2}+6}{x}$ and ${f}^{-1}(x)=1$ then find the value of $x$

If $2f(x)+3f(\frac{1}{x})=\frac{4{x}^{2}+6}{x}$ and ${f}^{-1}(x)=1$ then find the value of $x$

My Attempt:

$2f(x)+3f(\frac{1}{x})=\frac{4{x}^{2}+6}{x}$

$2f(x)+3f(\frac{1}{x})=4x+\frac{6}{x}$

At this point, I couldn't get other idea except comparing the corresponding terms. please provide any other method...

If $2f(x)+3f(\frac{1}{x})=\frac{4{x}^{2}+6}{x}$ and ${f}^{-1}(x)=1$ then find the value of $x$

My Attempt:

$2f(x)+3f(\frac{1}{x})=\frac{4{x}^{2}+6}{x}$

$2f(x)+3f(\frac{1}{x})=4x+\frac{6}{x}$

At this point, I couldn't get other idea except comparing the corresponding terms. please provide any other method...