5+6/2*9-50*2=
5+3*9-50*2
5+27-50*2
5+27+100
32+100
132
Hope this helps :)
From the given tabular data,
Volume of water (V) = 50.0 ml
Mass of water (m) = 49.7 g
So, the density of water (ρ) = Mass / Volume
or, = 49.7 g /50.0 ml = 0.994 g / ml
<u>Actual value of the density of water (ρ') = 0.998 g /ml</u>
So, the percentage error in the density of water
= [( actual value - observed value) / actual value ] ×100
or, = [ (0.998 g /ml - 0.994 g / ml ) / 0.998 g /ml ] ×100
or, = 0.4 %
Hence, the required % error in the density of water will 0.4 %.
Let's first denote the intersection point of diagonals with O. In this parallelogram, we can observe that ΔAOB is congruent to ΔCOD because of AB=CD and the angles ∠BAO and ∠OCD are equal (They are alternate angles). According to the rule of congruency, 3x=4x-5 and x=5
Based on the population mean, the mean of the distribution sample means would be 204.6. The standard deviation would be 0.22.
<h3>What are the mean and standard deviation of the distribution?</h3>
The mean of the distribution sample means is the same as the population means which is 204.6.
The standard deviation can be found as:
σₓ = σ / √n
= 3.1 / √197
= 0.22
Find out more on standard deviation at brainly.com/question/12402189.
#SPJ1
27-18=9 (difference between)
9/18=0.5 (9 as a percentage of original as a decimal)
0.5x100=50 (as a percentage)
50% increase
