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jenyasd209 [6]
3 years ago
14

−1.091 + 12.12 + (−1.1) lol heres another i have a few more

Mathematics
1 answer:
Natalka [10]3 years ago
3 0

Answer:

9.929

Step-by-step explanation:

-1.091 + 12.12 = 11.029

11.029+(-1.1) = 9.929

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solve y 6.3=0.9y thats all the question i dont understand it so thats why im trying to get the answer to the problem
guajiro [1.7K]
What it wants you to do is to put it into a form where on the one side of the equation there is y and nothing else. for this, we need to get rid of the 0.9.

how do we do that?

we divide both sides of the equation by 0.9:

6.3/0.9=y

that's ugly, right? but if both denominator and nominator are decimal fractions of the same sort; we can multiply both by 10 and get rid of the point:

63/09=y
and...
\frac{63}{9} =7


so basically 7=y, or y=7

and this is the solution!
8 0
3 years ago
If [infinity] cn8n n = 0 is convergent, can we conclude that each of the following series is convergent? (a) [infinity] cn(−3)n
pashok25 [27]

Answer:

a) we know that this is convergent.

b) we know that this might not converge.

Step-by-step explanation:

Given the \sum^\infty_{n=0}C_n8^n is convergent

Therefore,

(a)  \sum^\infty_{n=0}C_n(-3)^n The power series \sum C_nx^n has radius of convergence at least as big as 8. So we definitely know it converges for all x satisfying -8<x≤8. In particular for x = -3

∴ \sum^\infty_{n=0}C_n(-3)^n  is convergent.

(b) \sum^\infty_{n=0}C_n(-8)^n -8 could be right on the edge of the interval of convergence, and so might not converge

8 0
3 years ago
your plant runs two assembly line line a produces 427 units per hour and line b produces 519 units per hour how many more units
Juli2301 [7.4K]
Line b produces 92 more units per hour
3 0
3 years ago
Read 2 more answers
11. Simplify (6x^-2)^2 (0.5x)^4 Show your work
horrorfan [7]
Hope this helps you.

6 0
2 years ago
Read 2 more answers
A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.
nika2105 [10]

Answer:

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

\sigma=12 represent the population deviation

n=76 represent the sample size  

Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

Replacing we got:

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0
3 years ago
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