Answer:
396
Step-by-step explanation:
the volume of a rectangular is calculated as V=l×w×h
l=11, w=12, h=3
V=11×12×3
V= 396
Answer:
With the given margin of error its is possible that candidate A wins and candidate B loses, and it is also possible that candidate B wins and candidate A loses. Therefore, the poll cannot predict the winner and this is why race was too close to call a winner.
Step-by-step explanation:
A group conducted a poll of 2083 likely voters.
The results of poll indicate candidate A would receive 47% of the popular vote and and candidate B would receive 44% of the popular vote.
The margin of error was reported to be 3%
So we are given two proportions;
A = 47%
B = 44%
Margin of Error = 3%
The margin of error shows by how many percentage points the results can deviate from the real proportion.
Case I:
A = 47% + 3% = 50%
B = 44% - 3% = 41%
Candidate A wins
Case II:
A = 47% - 3% = 44%
B = 44% + 3% = 47%
Candidate B wins
As you can see, with the given margin of error its is possible that candidate A wins and candidate B loses, and it is also possible that candidate B wins and candidate A loses. Therefore, the poll cannot predict the winner and this is why race was too close to call a winner.
Answer:

Step-by-step explanation:
Given equation of ellipsoids,

The vector normal to the given equation of ellipsoid will be given by





Hence, the unit normal vector can be given by,



Hence, the unit vector normal to each point of the given ellipsoid surface is

C) 1059 skittles
Step-by-step explanation:
We know that the first container held 192 skittles. Knowing the dimension of the container we can calculate the volume of 192 skittles:
Volume of 192 skittles = 5 × 4 × 4 = 80 cm³
Volume of 1 skittle = 80 / 192 = 0.42 cm³
We also know that the second container held 258 skittles. Knowing the dimension of the container we can calculate the volume of 258 skittles:
Volume of 258 skittles = 12 × 3 × 3 = 108 cm³
Volume of 1 skittle = 108 / 258 = 0.42 cm³
We found that the volume of 1 skittle is equal to 0.42 cm³. Now we calculate the volume of the skittles jar:
volume of cylinder = π × radius² × height
volume of skittles jar = 3.14 × 3.5² × 11.5 = 442 cm³
Now we can calculate the number of skittles in the jar:
number of skittles in the jar = volume of the jar / skittle volume
number of skittles in the jar = 442 / 0.42 = 1052 which is close the C) 1059
Learn more about:
volume of cylinder
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Answer:
a = 540°
b= 180°
Step-by-step explanation:
a= pentagon has 5 sides which means 180 × 5= 540
b= 540÷5=180