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:
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
Step-by-step explanation:
Total plants = 11
Domestic plants = 7
Outside the US plants = 4
Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:
P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ
where n = total no. of trials
x = no. of successful trials
p = probability of success
q = probability of failure
Here we have n=4, p=4/11 and q=7/11
P(X≥1) = 1 - P(X<1)
= 1 - P(X=0)
= 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰
= 1 - 0.16399
P(X≥1) = 0.836
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
Answer:
Step-by-step explanation:
The estemit will be 0
<span>x^2+10x + 25 = (x + 5)^2
missing = 25
hope it helps</span>
