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2 years ago

The area of a circle is 95.75cm2. Find the length of the radius rounded to 2 DP.

Mathematics

1 answer:

notka56 [123]2 years ago

4 0

Answer:

5.52 cm

Step-by-step explanation:

area of circle =pi r^2

r=square root (95.75/pi)

=5.52

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How can you tell when an equation in one variable has infinitely many solutions or no solution?

Masteriza [31]

Answer:

A system has infinitely many solutions when it is consistent and the number of variables is more than the number of nonzero rows in the rref of the matrix. For example if the rref is ��has solution set (4-3z, 5+2z, z) where z can be any real number.

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2 years ago

Crop researchers plant 15 plots with a new variety of corn. The yields in bushels per acre are: 138.0 139.1 113.0 132.5 140.7 10

son4ous [18]

There is a relationship between confidence interval and standard deviation:
$\theta=\overline{x} \pm \frac{z\sigma}{\sqrt{n}}$
Where $\overline{x}$ is the mean, $\sigma$ is standard deviation, and n is number of data points.
Every confidence interval has associated z value. This can be found online.
We need to find the standard deviation first:
$\sigma=\sqrt{\frac{\sum(x-\overline{x})^2}{n}$
When we do all the calculations we find that:
$\overline{x}=123.8\\ \sigma=11.84$
Now we can find confidence intervals:
$($90\%,z=1.645): \theta=123.8 \pm \frac{1.645\cdot 11.84}{\sqrt{15}}=123.8 \pm5.0\\($95\%,z=1.960): \theta=123.8 \pm \frac{1.960\cdot 11.84}{\sqrt{15}}=123.8 \pm 5.99\\ ($99\%,z=2.576): \theta=123.8 \pm \frac{2.576\cdot 11.84}{\sqrt{15}}=123.8 \pm 7.87\\$
We can see that as confidence interval increases so does the error margin. Z values accociated with each confidence intreval also get bigger as confidence interval increases.
Here is the link to the spreadsheet with standard deviation calculation:
https://docs.google.com/spreadsheets/d/1pnsJIrM_lmQKAGRJvduiHzjg9mYvLgpsCqCoGYvR5Us/edit?usp=sharing

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3 years ago

A man that is 6 feet tall casts a shadow 5 feet long. At the same time, a lamppost beside the man casts a shadow 25 feet long. H

Montano1993 [528]

Answer:

The answer to your question is: L = 30 ft

Step-by-step explanation:

Data

man = 6 ft

man shadow = 5 ft

lamppost shadow = 25 ft

lamppost height = ? = L

Process

$\frac{6}{5} = \frac{L}{25}$

$L = \frac{6 x 25}{5}$

$L = \frac{150}{5}$

L = 30 ft

5 0

3 years ago

Please help. PLEEEEEEEEEEASSSSSSEEE<br> There are three right answers I don’t need and explanation

Angelina_Jolie [31]

Answer:

D,E,C

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$