Answer:
Step-by-step explanation:
5) 108
6) 55
7) 49
8) 65
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The volume of the cylindrical giant ocean tank is approximately 26,635.23 cubic feet.
The volume of a cylinder is defined as the space occupied by the cylinder and the volume of any three-dimensional shape is the space occupied by it.
The volume of the tank can be found by using the formula for the volume of a cylinder, which is V = πr²h, where r is the radius, h is the height (or depth in this case), and π is approximately 3.14.
Plugging in the given values, we get:
V = 3.14 × (18.8)² × 24
V = 3.14 × 353.44 × 24
V = 26,635.23 cubic feet
Thus, the volume of the cylindrical tank is approximately 26,635.23 cubic feet.
Learn more about the volume of the cylinder here :
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Answer:
Claudia could estimate the amount of water that her teammates drank by using this process:
1. Measure the amount that Claudia herself drank.
2. Subtract that from the amount of water that the jug was filled with before the soccer game.
3. Find the difference of that and the current amount of water.
Step-by-step explanation:
Answer:
Use a measuring cup to find the amount of water left in the jug, and then subtract the number of cups from 16.
Step-by-step explanation:
50:35
y=1/2x-5
Slope=
y-intercept=
What can she infer about the wingspans of the two types of birds?
Type 1: {18, 24, 20, 22, 26}
Type 2: {24, 21, 19, 26, 30}
A.
Type 1 and Type 2 birds have similar wingspan distributions.
B.
Type 1 and Type 2 birds have somewhat similar wingspan distributions.
C.
Type 1 birds and Type 2 birds do not have similar wingspan distributions.
D.
Type 1 birds and Type 2 birds have identical wingspan distributions.
Answer:
Step-by-step explanation:
The given data set for type 1 of birds is:
Type 1: {18, 24, 20, 22, 26}
Type 2: {24, 21, 19, 26, 30}
Mean of the type 1 data is:
Data
18 16
24 4
20 4
22 0
26 16
Now, mean average of squares is:
Standard deviation=
Now, the difference of mean and its standard deviation of type 1 data set is:
=22-2.828
Difference =19.172
The given data set for type 2 of birds is:
Type 2: {24, 21, 19, 26, 30}
Mean of the type 2 data is:
Data
24 0
21 9
19 25
26 4
30 36
Now, mean average of squares is:
Standard deviation=
Now, the difference of mean and its standard deviation of type 2 data set is:
=24-3.84
Difference=20.16
Since, the difference of mean and standard deviation of both type 1 and type 2 data set is different, therefore, Type 1 birds and Type 2 birds do not have similar wingspan distributions.
Hence, option C is correct.
Answer:
Type 1 and Type 2 birds have similar wingspan distributions.
Step-by-step explanation: