Given : - Square ABCD with side 3. E and F as midpoints.
To find : - area of EBFD
Solution : - We have, area of square ABCD = 3 x 3 = 9 units.
Thus, (ar)EBFD = ar ABCD - ar DAE - arDCF
arDAE = 1/2 x base x height
=1/2 x 1.5 x 3 ( AE is 1/2 of AB = 1.5, DA is altitude)
= 2.25
arDFC = 1/2 x base x height
= 1/2 x 1.5 x 3 (FC is 1/2 of BC, DC is altitude)
= 2.25
Thus, (ar) EBFD = arABCD - arDAE - arDCF
= 9 - 2.25 - 2.25
= 4.5 units.
Thus, area of quad EBFD is 4.5 units.
Answer:
42°
Step-by-step explanation:
For this problem, it may help you to draw it out to visualize it. You want to find the angle between the stream and road. Based on the description the triangle looks like this.
/ |
Stream(27) / | Road( 20)
/__|
Fence
You know the length of the road is 20, and the stream is 27. The angle can be found by using Cos∅=a/h. Adjacent to where the road and stream meet is the road, 20, and hypotenuse is the stream, 27.
So 20/27 = .74 = Cos∅
To find the angle ∅ use inverse cosine to get 42°.
Hope that helps and let me know if I did anything incorrectly!
Answer:
24
Step-by-step explanation:
The equations just need to be equal so you multiply 3 by 6 to get 8 and then 4 by 6 to get 24
3 / 4 = 18 / 24
Answer:
409.86
Step-by-step explanation:
230×.65=149.50
230+149.50=379.50×.08=30.36
379.50+30.36=409.86
Answer:
Step-by-step explanation:
The number of samples is large(greater than or equal to 30). According to the central limit theorem, as the sample size increases, the distribution tends towards normal. The formula is
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = population standard deviation
n = number of samples
From the information given,
µ = 22199
σ = 5300
n = 30
the probability that a senior owes a mean of more than $20,200 is expressed as
P(x > 20200)
Where x is a random variable representing the average credit card debt for college seniors.
For n = 30,
z = (20200 - 22199)/(5300/√30) =
- 2.07
Looking at the normal distribution table, the probability corresponding to the z score is 0.0197
P(x > 20200) = 0.0197
