Answer:
x=7
Step-by-step explanation:
m angle SAR = m angle SQR (angles lying in the same segment of a circle are always equal)
that's why
7x+7=9x-7
9x-7x= 7+7
2x= 14
x=7
Answer:
The probability distribution of the number of customers that enter the store on a given day is:
![P(x=k)=\frac{57^ke^{-57}}{k!}](https://tex.z-dn.net/?f=P%28x%3Dk%29%3D%5Cfrac%7B57%5Eke%5E%7B-57%7D%7D%7Bk%21%7D)
Step-by-step explanation:
To calculate a parameter for the given day, we have to calculate what is the average arrival rate for the day.
This can be done with the data given:
1) From 10 to 12, 8 arrival/hour: 16 expected arrivals in this period
2) From 12 to 2, 8 to 12 arrival/hour (average: 10 arrivals): 20 expected arrivals in this period.
3) From 2 to 5, from 10 to 4 arrival/hour (average: 7 arrivals): 21 expected arrivals in this period.
The total expected arrivals in a day are: 16+20+21=57 arrivals/day.
Then, the probability distribution of the number of customers that enter the store on a given day is:
![P(x=k)=\frac{57^ke^{-57}}{k!}](https://tex.z-dn.net/?f=P%28x%3Dk%29%3D%5Cfrac%7B57%5Eke%5E%7B-57%7D%7D%7Bk%21%7D)
Answer:
Mean qualifying speed
= 224.5669mi/hr
Speed that passed the average/mean
= 226.240mi/hr ,224.037mi/hr 226.484mi/hr, 225.172 mi/hr
Step-by-step explanation:
Let's understand what mean is then understand what mean qualifying speed is.
First mean is like the average of the seven speed listed below.
So let's find the mean.
Mean = (223.684+ 222.929+ 226.240
+224.037 +226.484+ 225.172
+223.422)/7
Mean = 1571.968/7
Mean = 224.5669
So the mean qualifying speed is those speed that qualifies when the mean is taken, i.e the speed that crossed the average.
Mean qualifying speed are
=226.240 ,224.037,226.484
, 225.172
2. QM is congruent to QN
4, All right angles are congruent
5.QX is congruent to QX
6.SAS congruence
7.MX is congruent to NX
Answer:
A 2√2(cos 7π/4 + i sin 7π/4)
Step-by-step explanation:
A. 2√2(cos 7π/4 + i sin 7π/4)
2 sqrt(2) ( sqrt(2)/2 - sqrt(2)/2 i)
Distribute
2-2i
This is in the fourth quadrant
B. 2√2(cos 150° + i sin 150°)
2 sqrt(2) (-sqrt(3)/2 +1/2i)
-sqrt(6) +sqrt(2) i
This is in the third quadrant (NO)
C. 2(cos 7π/4 + i sin 7π/4)
2( ( sqrt(2)/2 - sqrt(2)/2 i))
sqrt(2) - sqrt(2) i
This is the fourth quadrant
D. 2(cos 90° + i sin 90°)
2(0+i)
2i
This is on the positive y axis NO
Now we need to decide between the two in the fourth quadrant.
The point has an x coordinate of 2 and a y coordinate of -2
This aligns with point A