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lidiya [134]
3 years ago
15

if the value to the nearest thousandth, of cosO is -0.385, which of the following could be true about O? A. 0<O<pi/6 B. pi

/6<O<pi/3 C. pi/3<O<pi/2 D. pi/2<O<2pi/3 E. 2pi/3<O<pi
Mathematics
1 answer:
Dimas [21]3 years ago
4 0
Its the last choice there. Converting that to degrees you get the interval of 90<O<120, which is the only angle that lies in the quadrant in which cosine is negative, which is the second quadrant. Cosine is also negative in the third quadrant, but you don't have choices for the third quadrant.
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vodka [1.7K]

Answer:$259

Step-by-step explanation:

since we need to increase $250 by 6%, it is useful to find one percent because 1% is easy to find

to find one percent of 250, we move the decimal point(we cant see it right now) from after the number to after 2 giving us 2.50. we need to do this to find one percent because 1% is the same as one HUNDREDth, and since 100 has two zeros, we move the decimal place two spaces.now that we know that 1% is $2.50 ( which is the same as $2.5) we can times that by six which gives us 6%(because one times 6 is 6) to multiply 2.5 by 6, we first multiply 6 by 2 which will give us 12 and then multiply it by 0.5 (because 2 + 0.5 =2.5) 0.5 multiplied by 6 is equal to 3 and 3 + 6 = 9. we know know that 6% of $250 is $9. we now just add 9 to 250 leaving us with $259

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Answer:

-7

Step-by-step explanation:

(-\frac{3}{4}*12) + 2\\-\frac{36}{4} + 2\\-9 + 2\\-7

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Jan had 6 cookies more than Sam.Jan had 32 cookies .How many did Sam have?​
barxatty [35]

That means Sam has 26 cookies.

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A doctor administers a drug to a 37
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Answer:

Step-by-step explanation:

Daily dosage is 53 mg/kg(37 kg) = 1,961 mg

There are 24/4 = 6 four hour periods in each day.

So a four hour dosage should be 1961 / 6 =  327 mg

500 mg > 327 mg

So one tablet every 4 hours would exceed the recommended dosage by roughly 50%.

A closer dosage might be to take 1 tablet every 6 hours.

The suspension may be a better solution if effects of overdose are severe. A 5 mL dose every four hours would be very close to ideal.

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A fisherman catches fish according to a Poisson process with rate lambda = 0.6 per hour. The fisherman will keep fishing for two
spayn [35]

Answer:

Step-by-step explanation:

Given that a fisherman catches fish according to a Poisson process with rate lambda = 0.6 per hour.

The fisherman will keep fishing for two hours.

Since he continues till he gets atleast one fish, we can calculate probability as follows:

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= Prob (x=0) in I two hours and P(X≥1) in 3rd hour

=P(x=0)*P(X=0)*P(X≥1) (since each hour is independent of the other)

= 0.5488^2*(1-0.8781)\\\\=0.2645

(b) Find the probability that the total time he spends fishing is between two and five hours.

Prob that he does not get fish in I two hours * prob he gets fish between 3 and 5 hours

=P(0)^2 *F(1)^3\\=0.5488^2*0.2645^3\\=0.00557

(c) Find the expected number offish that he catches.

Expected value in Geometric distribution = \frac{1-p}{p}, where p = prob of getting 1 fish in one hour

= \frac{0.6}{1-0.6} \\=3

(d) Find the expected total fishing time, given that he has been fishing for four hours.

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= 1.25 hours

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