We know, 1 g = 1000 mg
so, 1 mg = 1/1000 g
then, 8,450 mg = 1/1000 * 8,450 = 8.450 g
In short, Your Answer would be: 8.450 Grams
Hope this helps!
Answer:
The nth term of the sequence is
<h2>5 + 2n</h2>
Step-by-step explanation:
The sequence above is an arithmetic sequence
For an nth term in an arithmetic sequence
A(n) = a + ( n - 1)d
where a is the first term
n is the number of terms
d is the common difference
From the question
a = 7
d = 9 - 7 = 2 or 11 - 9 = 2
So the nth term for the sequence is
A(n) = 7 + ( n - 1)2
= 7 + 2n - 2
<h3>A(n) = 5 + 2n</h3>
Hope this helps you
Answer:
854
Step-by-step explanation:
1000/15=66
1000-66-80=854
Answer:
Step-by-step explanation:
Given that a teacher gives a test to a large group of students. The results are closely approximated by a normal curve
mu =74 and sigma =8
A grade starts from 100-8 = 92nd percentile
Z score for 92nd percentile = 1.405
X score = 74+8(1.405) = 85.24
--------------------
B cut off is to next 16%
Hence C would start for scores below 100-(8+16) = 76%
76th percentile = 0.705*8+74 =79.64
Answer:
The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .
Step-by-step explanation:
Given as :
The mass of liquid water = 50 g
The initial temperature =
= 15°c
The final temperature =
= 100°c
The latent heat of vaporization of water = 2260.0 J/g
Let The amount of heat required to raise temperature = Q Joule
Now, From method
Heat = mass × latent heat × change in temperature
Or, Q = m × s × ΔT
or, Q = m × s × (
-
)
So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )
Or, Q = 50 g × 2260.0 J/g × 85°c
∴ Q = 9,605,000 joule
Or, Q = 9,605 × 10³ joule
Or, Q = 9605 kilo joule
Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer