The value of T in the equation if v = 42 is 12
Given:
T= v /6 + 5
If v = 42, find T
T= v /6 + 5
T = 42/6 + 5
T = 7 + 5
T = 12
Therefore, the value of T in the equation if v = 42 is 12
Learn more about algebra:
Positive
Step-by-step explanation:
A positive and a negative add up to a negative so if you add the sum of the first two then add the negative it will be positive
.
Answer:
92% Of confidence intervals to estimate the proportion of all fast-growing small companies that have a management succession plan.
(0.46154 , 0.558)
Step-by-step explanation:
Given sample size 'n' = 210
The sample proportion 'p' = 51% = 0.51
Confidence intervals are determined by
The 92% of z-score value
92% Of confidence intervals to estimate the proportion of all fast-growing small companies that have a management succession plan.
on calculation , we get
(0.51-0.048 , 0.51 + 0.048)
(0.46154 , 0.558)
Final answer:-
92% Of confidence intervals to estimate the proportion of all fast-growing small companies that have a management succession plan.
(0.46154 , 0.558)
Answer: just had this problem! X = 11
Answer:
7cm
Step-by-step explanation:
3cmx3cm=9cm
9cmx3.14=28.26
28.26/197.9=7 (plus a bunch of decimals!)
So 7cm is your height
Answer:
7.50 is the answer so either 7 or 8
Step-by-step explanation:
h= (A/2piR) -r
Plug in the area and the radius and you got the answer. Use 3.14 as pi
A) y = 96; The maximum number of moose that the forest can sustain at one time.
B) x = 96; The maximum number of moose that the forest can sustain at one time.
C) x = -1.6; The minimum number of moose that the forest can sustain at one time.
D) y = -1.6; The minimum number of moose that the forest can sustain at one time.
Answer:
V = π (-2 (ln 2)² + 4 ln 2 − 1)
V ≈ 2.55
Step-by-step explanation:
V = π ∫₁² (1 − (ln x)²) dx
V/π = ∫₁² (1 − (ln x)²) dx
V/π = ∫₁² dx − ∫₁² (ln x)² dx
V/π = x |₁² − ∫₁² (ln x)² dx
V/π = 1 − ∫₁² (ln x)² dx
To evaluate the second integral, integrate by parts.
If u = (ln x)², then du = 2 (ln x) / x dx.
If dv = dx, then v = x.
∫ u dv = uv − ∫ v du
= (ln x)² x − ∫ x (2 (ln x) / x) dx
= x (ln x)² − 2 ∫ ln x dx
Integrate by parts again.
If u = ln x, then du = 1/x dx.
If dv = dx, then v = x.
∫ u dv = uv − ∫ v du
= x ln x − ∫ x (1/x dx)
= x ln x − ∫ dx
= x ln x − x
Substitute:
∫ (ln x)² dx = x (ln x)² − 2 ∫ ln x dx
∫ (ln x)² dx = x (ln x)² − 2 (x ln x − x)
∫ (ln x)² dx = x (ln x)² − 2x ln x + 2x
Substitute again:
V/π = 1 − ∫₁² (ln x)² dx
V/π = 1 − (x (ln x)² − 2x ln x + 2x) |₁²
V/π = 1 + (-x (ln x)² + 2x ln x − 2x) |₁²
V/π = 1 + (-2 (ln 2)² + 4 ln 2 − 4) − (-1 (ln 1)² + 2 ln 1 − 2)
V/π = 1 − 2 (ln 2)² + 4 ln 2 − 4 + 2
V/π = -2 (ln 2)² + 4 ln 2 − 1
V = π (-2 (ln 2)² + 4 ln 2 − 1)
V ≈ 2.55