(n-1)+n+(n+1)
With n as the middle of 3 integers, n-1 is the integer just before n and n+1 is the integer just after n.
let's suppose x is the shortest leg of the triangle
The perimeter of the flower bed is 3x+8 ft
the surface to be coverd by sod would be x×(x+7)/2 sqft
(x+8)(x+8)=(x+7)^2+x^2
x'2+16x+64=x'2+14x+49+x^2
x^2-2x=15
x×(x-2)=15
x=5
(15+8)×$4.23=$97.29 for fencing
32.5sqft×$12.72=$413.4 for the sod
Answer/Step-by-step explanation:
Given:
C = right angle = 90°
BC = a = ?
AB = c = 12
AC = b = 9
Required:
a, <A, and <B
Solution:
✔️Find a using Pythagorean Theorem:
a = √(c² - b²)
a = √12² - 9²) = √63
a = 7.93725393 = 8 (nearest whole number)
✔️Find A by applying trigonometric ratio:
Thus,
Reference angle = A
Hypotenuse = 12
Adjacent = 9
Therefore,
m<A = 41° (nearest whole number)
✔️Find B:
m<B = 180 - (90 + 41) (sum of interior angles of triangle)
m<B = 49°
Answer:
- Two sided t-test ( d )
- 0.245782 ( c )
- Since P-value is too large we cannot conclude that the students’ weight are different for these two schools. ( c )
- The test is inconclusive; thus we cannot claim that the average weights are different. ( b )
Step-by-step explanation:
1) Test performed is a Two sided test and this because we are trying to determine the mean difference between two groups irrespective of their direction
<u>2) Determine the P-value ( we will use a data-data analysis approach on excel data sheet while assuming Unequal variances )</u>
yes No
Mean 94.47059 89.76471
Variance 173.2647 95.19118
Observations 17 17
df 30
t Stat 1.184211
P(T<=t) one-tail 0.122814
t Critical one-tail 1.697261
P(T<=t) two-tail 0.245782
Hence The p-value = 0.245782
3) Since P-value is too large we cannot conclude that the students’ weight are different for these two schools.
4) The test is inconclusive; thus we cannot claim that the average weights are different.
Answer:
A
Step-by-step explanation:
