To get the probability that at least 8 of the words on the test are words that students know we need to do the following calculations.
<span>=[C(16,8)*C(10,2) + C(16,9)*C(10,1) + C(16,10)*C(10,0)] / C(26,10)
</span>
=0.132
answer:
△JKI ≅ △CED
step-by-step explanation:
- like i mentioned before, they look similar and fulfill the SSS triangle theorem (this time's the theorem is different only)
Answer:
a) 3.128
b) Yes, it is an outerlier
Step-by-step explanation:
The standardized z-score for a particular sample can be determined via the following expression:
z_i = {x_i -\bar x}/{s}
Where;
\bar x = sample means
s = sample standard deviation
Given data:
the mean shipment thickness (\bar x) = 0.2731 mm
With the standardized deviation (s) = 0.000959 mm
The standardized z-score for a certain shipment with a diameter x_i= 0.2761 mm can be determined via the following previous expression
z_i = {x_i -\bar x}/{s}
z_i = {0.2761-0.2731}/{ 0.000959}
z_i = 3.128
b)
From the standardized z-score
If [z_i < 2]; it typically implies that the data is unusual
If [z_i > 2]; it means that the data value is an outerlier
However, since our z_i > 3 (I.e it is 3.128), we conclude that it is an outerlier.
Yes because 84+45 is 129 and 3(28+15) is 129
