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soldier1979 [14.2K]
3 years ago
7

Do) 6. The height of a cylinder is 5 centimeters. The circumference of the

Mathematics
1 answer:
nevsk [136]3 years ago
3 0

Answer:

2621.9cm3

Step-by-step explanation:

You might be interested in
Round off 787592799 to the nearest 10​
timama [110]

Answer:

787592800

Step-by-step explanation:

Hi there,

If you round this answer to the nearest tens place, it would be 787592800. Since the 799 is closer to 800 rather than 790, it would round up and change the last three digits of the number instead of the usual two digits.

Hope this answer helps. Cheers.

3 0
3 years ago
Can someone explain to me the steps how they got 2 √4^2+2x-3-2 from this question…
Naddik [55]

Step-by-step explanation:

with your square root symbol I never know what is inside the square root and what is possibly outside.

so, I can only guess and see what comes close.

f(x) = 2x² + x - 1

g(x) = sqrt(2x - 1) ??? is that so ?

h(x) = -2

2g(f(x)) + h(x)

g(f(x)) means that the whole f(x) expression is used as x in g(x).

the whole combined function is then

2×sqrt(2(2x² + x - 1) - 1) - 2

2×sqrt(4x² + 2x - 2 - 1) - 2

2×sqrt(4x² + 2x - 3) - 2

and if I am not mistaken, then this is the solution you mentioned at the beginning (if I try to read between the typos and missing info).

this is how people get to this.

do you understand it now ? or is there still something unclear ?

6 0
2 years ago
Estimate each product : <br> 1/7 x 20
Masteriza [31]

Answer:

2.85714285714

Step-by-step explanation:

just searched it.

5 0
3 years ago
Read 2 more answers
A random sample of 31 charge sales showed a sample standard deviation of $50. a 90% confidence interval estimate of the populati
Marysya12 [62]
The 100(1-\alpha)\% confidence interval of a standard deviation is given by:

\sqrt{ \frac{(n-1)s^2}{\chi^2_{1- \frac{\alpha}{2} } }} \leq\sigma\leq\sqrt{ \frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2} } }}

Given a sample size of 31 charge sales, the degree of freedom is 30 and for 90% confidence interval, 

\chi^2_{1- \frac{\alpha}{2} }=43.773 \\  \\ \chi^2_{\frac{\alpha}{2} }=18.493

Therefore, the 90% confidence interval for the standard deviation is given by

\sqrt{ \frac{(31-1)50^2}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(31-1)50^2}{18.493 }} \\  \\ \Rightarrow\sqrt{ \frac{(30)2500}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(30)2500}{18.493 }} \\  \\ \Rightarrow\sqrt{ \frac{75000}{43.773 }} \leq\sigma\leq\sqrt{ \frac{75000}{18.493 }} \\  \\ \Rightarrow \sqrt{1713.38} \leq\sigma\leq \sqrt{4055.59}  \\  \\ \Rightarrow41.4\leq\sigma\leq63.7
3 0
3 years ago
Need step by step help
MrRa [10]

Answer:

what do u think it is?

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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