Averages indicate 95 pounds per passenger with a standard deviation of 35 pounds, so multiplying this by the total passenger count of 200 gives the result of:
total average = 95 * 200 = 19,000
total std dev = 35 * 200 = 7,000
To find the z score:
z = (x – u) / s
here x represents the observed value = exceeding 20,000; u stands for the mean value = 19,000; s is the standard deviation = 7,000
z = (20,000 – 19,000) / 7,000
z = 0.143
Referring to distribution tables,
P (z = 0.14) = 0.5557
<span>Thus, there is a 55.57% likelihood that the total weight will surpass the 20,000-pound threshold</span>
Detailed derivation:
dA/dt = 6 - 0.02A
dA/dt = -0.02 (A - 300)
Rearranging terms.
dA / (A - 300) = -0.02 dt
Integrate both sides.
ln(A - 300) = -0.02t + C
Isolate A.
A - 300 = Ce^(-0.02t)
A = 300 + Ce^(-0.02t)
Apply initial condition to determine C.
50 = 300 + Ce^(-0.02 × 10)
50 = 300 + Ce^(-0.2)
-250 = Ce^(-0.2)
C = -250e^(0.2)
A = 300 - 250e^(0.2)e^(-0.02t)
A = 300 - 250e^(0.2 - 0.02t)
Answer: On average, a pet's weight during this vet visit is approximately 2.4 pounds away from 12.9 pounds.
If the MAD for another day's weights was 1.5, then that day’s weights would be less variable compared to the weights of pets seen today.
Step-by-step explanation:
Given: The data in the table outlines the weights of animals visiting a vet one day, in pounds.
Mean = 12.9
Median= 12.0
Mode = 12.0
Mean Absolute Deviation = 2.4
It’s understood that the mean absolute deviation (MAD) of a dataset indicates the average distance between each data point and the mean. It reflects the variation present in the dataset.
Therefore, the average weight of a pet visiting this vet on this day is about 2.4 pounds away from 12.9 pounds.
Moreover, if another day had a MAD of 1.5, and since 1.5 < 2.4,[TAG_42]]
it implies that the weights on that day would be less variable compared to those of pets encountered on this day.
Response:
a. 55 cars
b. 25 cars
Detailed explanation:
Let’s denote the quantity of cars with stereo systems as N(ss), those with air conditioning as N(ac), and those with sunroofs as N(sr).
We find that:
N(ss) = 30
N(ac) = 30
N(sr) = 40
N(ss and ac and sr) = 15
N(at least two) = 30
a.
To calculate how many cars possess at least one feature (N(at least one) or N(ss or ac or sr)), we apply:
N(ss or ac or sr) = N(ss) + N(ac) + N(sr) - N(ss and ac) - N(ss and sr) - N(ac and sr) + N(ss and ac and sr)
N(ss or ac or sr) = 30 + 30 + 40 - (N(at least two) + 2*N(ss and ac and sr)) + 15
Substituting, we find N(ss or ac or sr) = 30 + 30 + 40 - (30 + 2*15) + 15 = 55
b.
For those cars that have exactly one feature, we have:
N(only one) = N(at least one) - N(at least two)
N(only one) = 55 - 30 = 25
Answer:
You need to multiply the denorminator both sides in order to form x note the subject:
