Answer:
Hence the probability of the at least 9 of 10 in working condition is 0.3630492
Step-by-step explanation:
Given:
total transistors=100
defective=20
To Find:
P(X≥9)=P(X=9)+P(X=10)
Solution:
There are 20 defective and 80 working transistors.
Probability of at least 9 of 10 should be working out 80 working transistors
is given by,
P(X≥9)=P(X=9)+P(X=10)
<em>{80C9 gives set of working transistor and 20C1 gives 20 defective transistor and 100C10 is combination of shipment of 10 transistors}</em>
P(X≥9)=
<em>(Use the permutation and combination calculator)</em>
P(X≥9)=(231900297200*20/17310309456440)
+(1646492110120/17310309456440)
P(X≥9)=0.267933+0.0951162
P(X≥9)=0.3630492
Answer:
x =41.18114952
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
We know the hypotenuse and adjacent sides
cos theta = adjacent / hypotenuse
cos x = 14.3/19
taking the inverse cos of each side
cos^-1 (cos x) = cos ^-1 ( 14.3/19)
x =41.18114952
Answer:
c is the right answer of this
First, find out how much ribbon there is.
3 x 12 = 36
Now, see how many times you can divde that by 8.
36 can only be divided by 8 four times, and there would be 4 inches of ribbon left.
So, you would be able to make 4 bows, with 4 inches of ribbon remaining.