There are 60 seconds in a minute.
There are 60 minutes in an hour.
This means that there'd be 600 minutes in 10 hours.
And if there are 60 seconds in a minute, there are 600 x 60 seconds in 600 minutes.
600 x 60 = 600 x 10 x 6 = 6000 x 6 = 36000
So we know that there are 36000 seconds in 10 hours.
Answer:
One solution.
Step-by-step explanation:
To determine the number of possible solutions for a triangle with A = 113° , a = 15, and b = 8, we're going to use the law of sines which states that: "<em>When we divide side a by the sine of angle A it is equal to side b divided by the sine of angle B, and also equal to side c divided by the sine of angle C</em>".
Using the law of sines we have:


Solving for B, we have:

∠B = 29.4°
Therefore, the measure of the third angle is: ∠C = 37.6°
There is another angle whose sine is 0.4909 which is 180° - 29.4° = 150.6 degrees. Given that the sum of all three angles of any triangle must be equal to 180 deg, we can't have a triangle with angle B=113° and C=150.6°, because B+C>180.
Therefore, there is one triangle that satisfies the conditions.
Answer:
144
Step-by-step explanation:
Answer:
The confidence interval is (24116.3878,24883.6122).
Step-by-step explanation:
We are given the following information in the question:
Population mean,
= $23,500
Sample mean,
= $24,500
Sample standard deviation,s = $2,800
Sample size, n = 146
Confidence interval:
Putting the values, we get,
The confidence interval is (24116.3878,24883.6122).
Answer: just turn them into decimals and use a calculator then turn them back into fractions in simplified form
Step-by-step explanation: