In a 40-day period, at least one bear was observed by a camper in a designated camp site on 28 separate days. according to this information, what is the daily frequency of bear sightings (in which at least one bear is a.143%. b.70%. c.28%. d. 40%

So the solution we're looking for would be (28 days∗100) / 40 days = 70%

The final answer is B. 70%.

B. 70%

Step-by-step explanation:

Have a nice day :).

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐