Find tan(a) in the triangle Any help will be appreciated thank you
1 answer:
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<h2><em>Answer:</em></h2><h2><em>4</em><em>2</em><em> </em><em>cm^</em><em>2</em></h2>
<em>please </em><em>see</em><em> the</em><em> attached</em><em> picture</em><em> for</em><em> full</em><em> solution</em>
<em>hope </em><em>it</em><em> helps</em><em>.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em>
<span>1.
P(</span>at least one of the calculators is defective)=
1- P(none of the selected calculators is defective).
2.
P(none of the selected calculators is defective)
=n(ways of selecting 4 non-defective calculators)/n(total selections of 4)
3.
selecting 4 non-defective calculators can be done in C(35, 4) many ways,
where
while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,
4. so, P(none of the selected calculators is defective)=
5. P(at least one of the calculators is defective)=
1- P(none of the selected calculators is defective)=1-0.18=0.82
Answer:0.82
P(</span>at least one of the calculators is defective)=
1- P(none of the selected calculators is defective).
2.
P(none of the selected calculators is defective)
=n(ways of selecting 4 non-defective calculators)/n(total selections of 4)
3.
selecting 4 non-defective calculators can be done in C(35, 4) many ways,
where
while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,
4. so, P(none of the selected calculators is defective)=
5. P(at least one of the calculators is defective)=
1- P(none of the selected calculators is defective)=1-0.18=0.82
Answer:0.82