Answers in Scientific Notation.
1 answer:
one way is to factor the expressions
(a)
(1.2 + (5.35 ×
))
= (1.2 + 53.5)
= 54.7 × = 5.47 ×
×
= 5.47 ×
(b)
((6.91 ×
) + 2.4)
= (69.1 + 2.4)
= 71.5 × = 7.15 ×
×
= 7.15 ×
(c)
((9.7 ×
) + 8.3)
= (97 + 8.3)
= 105.3 × = 1.053 ×
×
= 1.053 ×
(d)
((3.67 ×
) - 1.6)
= (36.7 - 1.6)
= 35.1 × = 3.51 ×
×
= 3.51 ×
(e)
((8.41 ×
) - 7.9)
= (84.1 - 7.9 )
= 76.2 × = 7.62 ×
×
= 7.62 ×
(f)
((1.33 ×
) - 4.9 )
= (13.3 - 4.9) = 8.4 ×
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Answer:
Probability that fewer than 2 of these parts are defective is 0.604.
Step-by-step explanation:
We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.
A random sample of 7 parts produced by this machine is chosen.
The above situation can be represented through Binomial distribution;
where, n = number of trials (samples) taken = 7 parts
r = number of success = fewer than 2
p = probability of success which in our question is % of defective
parts produced by one of the machine, i.e; 19%
<em>LET X = Number of parts that are defective</em>
<u>So, it means X ~ Binom(n = 7, p = 0.19)</u>
Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)
P(X < 2) = P(X = 0) + P(X = 1)
=
=
= <u>0.604</u>
<em>Therefore, the probability that fewer than 2 of these parts are defective is 0.604.</em>