Answer:
m(∠AOF) = 148°
Step-by-step explanation:
From the figure attached,
CD intersects line EF at a point O.
Line CD is perpendicular to the line EF.
m(∠AOE) = 32°
m(∠COE) = 90°
Since m(∠COE) = m(∠AOE) + m(∠AOC) = 90°
32° + m(∠AOC) = 90°
m(∠AOC) = 90° - 32° = 58°
m(∠AOF) = m(∠AOC) + m(∠COF)
= 58° + 90°
= 148°
Therefore, m(∠AOF) = 148° will be the answer.
A, use the Pythagorean theorem. A squared plus B squared equals C Squared, or A+B=C. In other words, 24 (Which is C) - 8 (Which is B)=16 (Which is A).
Make use of the fact that side lengths in similar triangles are proportional.
.. 30/18 = (20 +x)/20
.. 20*(30/18) -20 = x . . . . multiply by 20, subtract 20
.. 33 1/3 -20 = x . . . . . . . . evaluate the product
.. 13 1/3 = x
Answer:
Step-by-step explanation:
Hello!
a)
The given information is displayed in a frequency table, since the variable of interest "height of a student" is a continuous quantitative variable the possible values of height are arranged in class intervals.
To calculate the mean for data organized in this type of table you have to use the following formula:
X[bar]= (∑x'fi)/n
Where
x' represents the class mark of each class interval and is calculated as (Upper bond + Lower bond)/2
fi represents the observed frequency for each class
n is the total of observations, you can calculate it as ∑fi
<u>Class marks:</u>
x₁'= (120+124)/2= 122
x₂'= (124+128)/2= 126
x₃'= (128+132)/2= 130
x₄'= (132+136)/2= 134
x₅'= (136+140)/2= 138
Note: all class marks are always within the bonds of its class interval, and their difference is equal to the amplitude of the intervals.
n= 7 + 8 + 13 + 9 + 3= 40
X[bar]= (∑x'fi)/n= [(x₁'*f₁)+(x₂'*f₂)+(x₃'*f₃)+(x₄'*f₄)+(x₅'*f₅)]/n) = [(122*7)+(126*8)+(130*13)+(134*9)+(138*3)]/40= 129.3
The estimated average height is 129.3cm
b)
This average value is estimated because it wasn't calculated using the exact data measured from the 40 students.
The measurements are arranged in class intervals, so you know, for example, that 7 of the students measured sized between 120 and 124 cm (and so on with the rest of the intervals), but you do not know what values those measurements and thus estimated a mean value within the interval to calculate the mean of the sample.
I hope this helps!
Answer:
48
Step-by-step explanation:
