Answer:
I believe the answer is y = -8
36-9
-9+36
1 (36-9)
36-(3×3)
Answer:
0.9995
Step-by-step explanation:
10% = 0.10
1 - 0.10 = 0.9
n = number of light bulbs = 7
we calculate this using binomial distribution.
p(x) = nCx × p^x(1-p)^n-x
our question says at most 4 is defective
= (7C0 × 0.1⁰ × 0.9⁷) + (7C1 × 0.1¹ × 0.9⁶) + (7C2 × 0.1² × 0.9⁵) + (7C3 × 0.1³ × 0.9⁴) + (7C4 × 0.1⁴ × 0.9³)
= 0.478 + 0.372 + 0.1239 + 0.023 + 0.0026
= 0.9995
we have 0.9995 probability that at most 4 light bulbs are defective.
7.32 , 13.1
you add
(1)0.92
+6.4
--------
7.32
now second operation subtract
15.74
- 2.64
--------
13 .10
