Find the value of x in the triangle below. Please help! :) I’ll give brainliest
2 answers:
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Answer:
Hence the correct option is option b) Yes, the upper level of one standard deviation is 72.
A score of 74 is not within one standard deviation of the mean.
Step-by-step explanation:
Here the given details are,
Mean = 68
SD = 4
Distribution is normal.
Z-score for x = 74 is given as below:
So, the score of 74 is 1.5 standard deviations from the mean.
Therefore the score is not lies between 64 and 72.
Yes, the upper level of one standard deviation is 72.
Check the picture below.
that's done just with the provided links, and I simply ran a line, so it'll be thereabouts, roughly between 13 and 14, depending on how you hang the line.
anyhow, if you have a TI(texas instruments) 89+ or higher, you can just provide the value pairs of the data and run a quadratic regression, or an exponential.
there are some online regression calculators as well.
that's done just with the provided links, and I simply ran a line, so it'll be thereabouts, roughly between 13 and 14, depending on how you hang the line.
anyhow, if you have a TI(texas instruments) 89+ or higher, you can just provide the value pairs of the data and run a quadratic regression, or an exponential.
there are some online regression calculators as well.
Answer:
a. b. 7.44 s
Step-by-step explanation:
a. If the concentration of salt in the brine entering the tank is 0.2 kg/L, determine the mass of salt in the tank after t min.
Let m(t) be the mass of salt in the tank at any time, t.
Now, since a brine solution flows in at a rate of 4 L/min and has a concentration of 0.2 kg/L, the mass flowing in per minute is m' = 4 L/min × 0.2 kg/L = 0.8 kg/min
Now, the concentration in the tank of volume 100 L at any time, t is m(t)/100 L. Since water flows out at a rate of 3 L/min, the mass flowing out per minute is
m(t)/100 × 3 L/min = 3m(t)/100 kg/min
Now the net rate of change of mass of salt in the tank per minute dm/dt = mass flowing in -mass flowing out
dm/dt = 0.8 kg/min - 3m(t)/100 kg/min
So, dm/dt = 0.8 - 0.03m(t)
The initial mass of salt entering m(0) = 0 kg
dm/dt = 0.8 - 0.03m(t)
separating the variables, we have
dm/[0.8 - 0.03m(t)] = dt
Integrating, we have
∫dm/[0.8 - 0.03m(t)] = ∫dt
-0.03/-0.03 × ∫dm/[0.8 - 0.03m(t)] = ∫dt
1/(-0.03)∫-0.03dm/[0.8 - 0.03m(t)] = ∫dt
-1/0.03㏑[0.8 - 0.03m(t)] = t + C
㏑[0.8 - 0.03m(t)] = -0.03t - 0.03C
㏑[0.8 - 0.03m(t)] = -0.03t + C' (C'= -0.03C)
taking exponents of both sides, we have
So, the mass of the salt after t min is
b. When will the concentration of salt in the tank reach 0.1 kg/L?
When the concentration of the salt reaches 0.1 kg/L, m(t) = 0.1 kg/L
Solving the equation for t,
taking natural logarithm of both sides, we have
-0.03t = ㏑0.9963
-0.03t = -0.0038
t = -0.0038/-0.03
t = 0.124 min
t = 0.124 × 60 s
t = 7.44 s