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In-s [12.5K]
3 years ago
12

Recess began at 11:12am. It ended at 11:44am. How long was recess?

Mathematics
2 answers:
Soloha48 [4]3 years ago
8 0

Answer:

32 minutes

Step-by-step explanation:

44-12=32

kolezko [41]3 years ago
8 0

Answer:

32 minutes

Step-by-step explanation:

Both times were at 11 am and some minutes. The difference is only in the minutes.

44 - 12 = 32

Answer: Recess was 32 minutes long.

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8 0
3 years ago
Car is driving at 30 kilometers per hour. How far in meters does it travel in 3 seconds?
Alenkasestr [34]

Answer:

25 meters

Step-by-step explanation:

30 km/hr = (30 km/1 hr)*(1000 m/1 km)*(1 hr/60 min)*(1 min/60 sec)

30 km/hr = (30*1000*1*1)/(1*1*60*60) meters per second

30 km/hr = 8.3333333333 m/s (approximate)

----------

The speed or rate is r = 8.3333333333 meters per second approximately, and the time is t = 3 seconds, so,

distance = rate*time

d = r*t

d = 8.3333333333 * 3

d = 24.9999999999

Due to rounding error, the true answer should be d = 25. We can see this if we opted to use the fraction from of 8.3333... instead of the decimal representation which isn't a perfect exact measurement.

4 0
3 years ago
How can you decide one funny a common multiple is useful in solving a question
Alisiya [41]
What? I have an idea of what your saying but i'm not positive. Can you explain better please?
8 0
3 years ago
The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =
gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is \mathbf{P(X>1100n) = 0.158655}

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy  is given as :

f_x(x)  = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0

Thus, the expected  total claim amount \mu =  1000

The variance of the total claim amount \sigma ^2  = 1000^2

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})

P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})

P(X>1100n) = P(Z> \dfrac{10*100}{1000})

P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345

\mathbf{P(X>1100n) = 0.158655}

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is \mathbf{P(X>1100n) = 0.158655}

4 0
3 years ago
Using the graph,order the lines from steepest slope to the least steep slope.
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Answer:

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3 years ago
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