The least number of times that two planes can cross is zero, since parallel planes do not intersect. A good example would be a floor and a ceiling, which run parallel, hence they do not meet. Conversely, if two planes occupy the same space, they can intersect at infinitely many points, as could happen with a line within that plane.
Answer:
C. The mean daily salary is not above $350 per day
Step-by-step explanation:
The calculation is displayed below:
Y = a + bX
where,
Y = earnings of a randomly chosen agent
a = $150
b = $50
X = number of loans closed
Now
E(x) = (1 × 0.05) + (2 × 0.10) + (3 × 0.22) + (4 × 0.30) + (5 × 0.18) + (6 × 0.12) + (7 × 0.03)
= 3.94
Consequently,
E(y) = $150 + ($50 × 3.94)
= $347
therefore, option C is incorrect
The given road capacity is 3,500 vehicles per hour, and the expected number of vehicles arriving is 14,000. To calculate the time required for these vehicles to access the arena:
14,000 vehicles divided by 3,500 vehicles per hour equals 4 hours.
If the event is scheduled to commence at 7 p.m., the roads should close at: 7 p.m. minus 4 hours, which results in 3 p.m.
Solution:
In Mr. Skinner's class, the count of students bringing lunch from home is 12 out of 20.
Fraction of students who brought lunch from home in Mr. Skinner's class=
For Ms. Cho's class, the number who brought lunch from home is 14 out of 21.
Fraction of students who brought lunch from home in Ms. Cho's class=
Siloni is utilizing two spinners with 15 equal sections to randomly select students from the classes and predict whether they brought lunch or will purchase it from the cafeteria.
Number of Equal sections in each Spinner=15
To visualize the students from Mr. Skinner's class who brought lunch using a Spinner with 15 equal sections =
For Ms. Cho's class, using a Spinner with 15 equal sections =
Mr. Skinner's Class +1 = Ms. Cho's Class
This means that the spinner for Ms. Cho's class will include one additional section representing students who brought lunch.
Option A signifies that one additional section on Mr. Skinner's spinner represents students who brought lunch, reflecting Ms. Cho's class.
Answer:
Approximately 59 stacked cups.
Step-by-step explanation:
We have the following measurements:
Height of one cup = 12.5 cm,
Height of two cups stacked = 14 cm,
Height of three cups stacked = 15.5 cm,
...and so on.
This situation can be described by an arithmetic sequence,
12.5, 14, 15.5,....
The first term is defined as a = 12.5,
with a common difference of d = 1.5 cm.
Thus, the height of x stacked cups is given by

As per the problem,
h(x) = 200
⇒ 1.5x + 11 = 200
⇒ 1.5x = 189
⇒ x = 59.3333333333 ≈ 59.
Therefore, you will need approximately 59 stacked cups.