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True [87]
3 years ago
15

12

Mathematics
1 answer:
Elan Coil [88]3 years ago
6 0
I think it’s 12xt1=13 y
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The LCM of 12 and 25 is 300
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Solve the system: x' = (2 4 -1 6)x
sertanlavr [38]

Answer:

x=e^{8t}+c

Step-by-step explanation:

we have given x'=(24-16)x=8x

x' denotes the differenation

\frac{dx}{dt}=8x  differentiation is performed with respect to t

by variable separable method we can write  

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6 0
4 years ago
7 measured data points have a sample mean of 1403 and a standard deviation of 27. Determine the best estimate of the mean value
mina [271]

Answer:

Step-by-step explanation:

Hello!

You have sample of n=7 with mean X[bar]= 1403 and standard deviation S=27 and are required to estimate the mean with a 95%CI.

Asuming this sample comes from a normal population I'll use a stuent t to estimate the interval (a sample of 7 units is too small for the standard normal to be accurate for the estimation):

[X[bar]±t_{n-1;1-\alpha /2} * \frac{S}{\sqrt{n} }]

t_{n-1;1-\alpha /2} = t_{6;0.975}= 2.365

[1403±2.365*\frac{27}{\sqrt{7} }]

[1378.87;1427.13]

The margin of error is the semiamplitude of the interval and you can calculate it as:

d= \frac{Upbond-Lowbond}{2}= \frac{1427.13-1378.87}{2} = 24.13

With a confidence level of 95% you'd expect that the real value of the mean is contained by the interval [1378.87;1427.13], the best estimate of the mean value is expected to be ± 24.13 of 1403.

I hope it helps!

4 0
3 years ago
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