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3 years ago

6

Would like to know what I am doing wrong and have an answer

1 answer:

Paul [167]3 years ago

4 0

A can is a cylinder shape.

Volume of a cylinder = PI x r^2 X h

r is the radius and h is the height.

The diameter is given as 3.8, the radius is half that: 3.8 /2 = 1.9

Height is given as 6.5

Volume = 3.14 x 1.9^2 x 6.

Volume = 73.6801 cubic inches.

Rounded to the nearest tenth = 73.7 cubic inches.

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The midpoint of CD is M(-1,-2). One endpoint is C (-9,-4). Find the coordinates of the other endpoint D.

yan [13]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&C&(~ -9 &,& -4~) 
% (c,d)
&D&(~ x &,& y~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{x-9}{2}~~,~~\cfrac{y-4}{2} \right)=\stackrel{midpoint~M}{(-1~~,~~-2)}\implies 
\begin{cases}
\cfrac{x-9}{2}=-1\\\\
x-9=-2\\
\boxed{x=7}\\
----------\\
\cfrac{y-4}{2}=-2\\\\
y-4=-4\\
\boxed{y=0}
\end{cases}$

6 0

3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago

Plz help me with this two question

alekssr [168]

Part A:

3.14 X (4)^2 X 3/3
= 50.26548246

5 0

2 years ago

Read 2 more answers

PLZ HELP!!!! Predict the number of times you will win a prize in 500 spins, based on the theoretical probability.

Over [174]

I think the answer is B. 150 i may be wrong though

3 0

3 years ago

Read 2 more answers

Shannon walks 2,500 metres in 1 hour 40 minutes. <br> How far does she walk in one minute?

stich3 [128]

Answer:

25m

Step-by-step explanation:

Shannon's rate is 2,500m/100min. Simplifying, we see that this is equal to 25m/1min.

So, she walks 25 m in one minute.

4 0

3 years ago