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Anettt [7]
2 years ago
13

What is the volume of a cylinder with a radius of 9 inches and a height of 2 inches​

Mathematics
1 answer:
White raven [17]2 years ago
4 0

Answer:

508.94

Step-by-step explanation:

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On the coloring sheet, thd a ratio<br> is equivalent to 3/10.
DerKrebs [107]

Answer:

Equivalent ratios:

3 : 10   6 : 20   9 : 30   12 : 40   15 : 50   18 : 60   21 : 70   24 : 80   27 : 90   30 : 100   33 : 110   36 : 120   39 : 130   42 : 140   45 : 150   48 : 160   51 : 170   54 : 180   57 : 190   60 : 200   63 : 210   66 : 220   69 : 230   72 : 240   75 : 250   78 : 260   81 : 270   84 : 280   87 : 290   90 : 300

<u>( Brainlyst will help my rank <3 )</u>

8 0
2 years ago
How would i do this problem <br>slope= -2, y-intercept = 7
marin [14]
SINCE the slope is -2 it would be like this -2/1 but first plot the y-intercept onto the graph and go down 2 and over 1.
4 0
2 years ago
Read 2 more answers
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
2 years ago
I really need help. ​
fiasKO [112]
39/10 I’m not sure! Use the app called photo math
8 0
3 years ago
Need help, cant figure it out
just olya [345]

Answer:

<WSN=80

Step-by-step explanation:

arc WN = 160

<wsn= 1/2 of arc WN

7 0
3 years ago
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