Answer:
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Answer:
ρ_air = 0.15544 kg/m^3
Step-by-step explanation:
Solution:-
- The deflated ball ( no air ) initially weighs:
m1 = 0.615 kg
- The air is pumped into the ball and weight again. The new reading of the ball's weight is:
m2 = 0.624 kg
- The amount of air ( mass of air ) pumped into the ball can be determined from simple arithmetic between inflated and deflated weights of the ball.
m_air = Δm = m2 - m1
m_air = 0.624 - 0.615
m_air = 0.009 kg
- We are to assume that the inflated ball takes a shape of a perfect sphere with radius r = 0.24 m. The volume of the inflated ( air filled ) ball can be determined using the volume of sphere formula:
V_air = 4*π*r^3 / 3
V_air = 4*π*0.24^3 / 3
V_air = 0.05790 m^3
- The density of air ( ρ_air ) is the ratio of mass of air and the volume occupied by air. Expressed as follows:
ρ_air = m_air / V_air
ρ_air = 0.009 / 0.05790
Answer: ρ_air = 0.15544 kg/m^3
Answer:
D
Step-by-step explanation:
literlay just count
Variance is the standard deviation squared but we're not going to use that now. Let's first calculate the mean:
mean = (17+13+13+22+11+20)/6 = 16.
Now for each value, let's see how far it is from this mean. We'll square these distances and average them. That's our variance.
17 distance 1 squared = 1
13 distance 3 squared = 9
13 distance 3 squared = 9
22 distance 6 squared = 36
11 distance 5 squared = 25
20 distance 4 squared = 16
Now average these outcomes:
variance = (1+9+9+36+25+16)/6 = 16.
So the variance by coincidence is the same as the mean.
Answer C is your answer.
384 step by step explanation:
