<span><span>27100</span>(60)=<span>815</span>=16<span>15</span></span>
The 16 is how many minutes you have. If you were not given the seconds, we would have to say that we have a remainder of (1/5) of a minute. Since 1 minute is (1/60), one-fifth of that is (1/300). Now we would need to see how many seconds that gives us. 1 second is (1/3600) of a degree, so we would need to divide (1/300) by (1/3600). Doing this gives us:
<span><span><span>1300</span>÷<span>13600</span>=<span>1300</span>(3600)=12</span></span>
Answer:
H0: μd=0 Ha: μd≠0
t= 0.07607
On the basis of this we conclude that the mean weight differs between the two balances.
Step-by-step explanation:
The null and alternative hypotheses as
H0: μd=0 Ha: μd≠0
Significance level is set at ∝= 0.05
The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571
The test statistic under H0 is
t = d/ sd/ √n
Which has t distribution with n-1 degrees of freedom
Specimen A B d = a - b d²
1 13.76 13.74 0.02 0.004
2 12.47 12.45 0.02 0.004
3 10.09 10.08 0.01 0.001
4 8.91 8.92 -0.01 0.001
5 13.57 13.54 0.03 0.009
<u>6 12.74 12.75 -0.01 0.001</u>
<u>∑ 0.06 0.0173</u>
d`= ∑d/n= 0.006/6= 0.001
sd²= 1/6( 0.0173- 0.006²/6) = 1/6 ( 0.017294) = 0.002882
sd= 0.05368
t= 0.001/ 0.05368/ √6
t= 0.18629/2.449
t= 0.07607
Since the calculated value of t= 0.07607 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the mean weight differs between the two balances.
Answer:
Option C = 335.5 in²
Step-by-step explanation:
Area of triangle = (1/2)×b×h
= (1/2)×a×a
= (1/2)×11×11
=60.5 in²
Area of rectangle = l×b
= b×a
= 25×11
= 275 in²
Area of pencil = Area of triangle + Area of rectangle
=60.5 + 275
335.5 in²
Answer:
1 / 962598
Step-by-step explanation:
Let S be the sample space
total number of possible outcomes = n(S)
Let E be the event
total number of favorable outcomes = n(E)
Compute the number of ways to select 5 numbers from 0 through 42:
Total numbers to choose from = 43
So
Total number of ways to select 5 numbers from 43
= n(S) = 43C5
= 43! / 5! ( 43-5)!
= 43! / 5! 38!
= 43*42*41*40*39*38! / (5*4*3*2*1)*38!
= 115511760/120
n(S) = 962598
Hence there are 962598 ways to select 5 numbers from 43
Compute the probability of being a Big Winner
In order to be a Big Winner all 5 of the 5 winning balls are to be chosen and there is only one way you can for this event to occur. So
n(E) = 1
Here E is to be a Big Winner
So probability of being a Big Winner = P(E)
= n(E) / n(S)
= 1 / 962598
Hence
P(being a Big Winner) = P(E) = 1 / 962598
