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AveGali [126]
3 years ago
13

A pair of events are observed to have coordinates (0s,0m) and (50.0s,9.00×109m) in a frame s. what is the proper time interval δ

τ between the two events?
Mathematics
1 answer:
Ludmilka [50]3 years ago
5 0
To find out time interval δt we need to substract initial time from the final time. In this question first number in the coordinates representes time:

δt=50 - 0
δt= 50s

TIme interval is 50s.
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(a). Consider the parabolic function f(x) = ax² +bx+c, where a ±0, b and care constants. For what values of a, band c is f
ki77a [65]

Information about concavity is contained in the second derivative of a function. Given f(x) = ax² + bx + c, we have

f'(x) = 2ax + b

and

f''(x) = 2a

Concavity changes at a function's inflection points, which can occur wherever the second derivative is zero or undefined. In this case, since a ≠ 0, the function's concavity is uniform over its entire domain.

(i) f is concave up when f'' > 0, which occurs when a > 0.

(ii) f is concave down when f'' < 0, and this is the case if a < 0.

In Mathematica, define f by entering

f[x_] := a*x^2 + b*x + c

Then solve for intervals over which the second derivative is positive or negative, respectively, using

Reduce[f''[x] > 0, x]

Reduce[f''[x] < 0, x]

5 0
3 years ago
Can somebody please help me I really don’t understand this??
Viefleur [7K]

Answer:

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7 0
3 years ago
A test was marked out of 80. Aboy scored
GREYUIT [131]

Answer:

B

Step-by-step explanation:

To solve this you do 80/100=.8

You than do .8×60= 48

6 0
4 years ago
Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 in
nadezda [96]

Answer:

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:

X \sim N(288,3.7)  

Where \mu=288 and \sigma=3.7

The confidence level is 95.44 and the signficance is 1-0.9544=0.0456 and the value of \alpha/2 =0.0228. And the critical value for this case is z = \pm 1.99

Using this condition we can find the limits

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

8 0
3 years ago
Pls I need answer it’s due today!!!!!!!!!
vesna_86 [32]

Step-by-step explanation:

in probability,or means addition.

14/28+19/28

1/2+19/28

you then find the LCM of the denominator.

14+19/28

33/28

you then divide 33 by 28.

6 0
3 years ago
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