Let y = √4+7t
then u= 4+7t
y=√u = u^½
du/dt= 7
dy/du = ½U^-½
dy/dt = du/dt • dy/du
= 7×½U^-½
= 7/2√U
= 7 / (2{√4+7t})
Answer:
A) 68.33%
B) (234, 298)
Step-by-step explanation:
We have that the mean is 266 days (m) and the standard deviation is 16 days (sd), so we are asked:
A. P (250 x < 282)
P ((x1 - m) / sd < x < (x2 - m) / sd)
P ((250 - 266) / 16 < x < (282 - 266) / 16)
P (- 1 < z < 1)
P (z < 1) - P (-1 < z)
If we look in the normal distribution table we have to:
P (-1 < z) = 0.1587
P (z < 1) = 0.8413
replacing
0.8413 - 0.1587 = 0.6833
The percentage of pregnancies last between 250 and 282 days is 68.33%
B. We apply the experimental formula of 68-95-99.7
For middle 95% it is:
(m - 2 * sd, m + 2 * sd)
Thus,
m - 2 * sd <x <m + 2 * sd
we replace
266 - 2 * 16 <x <266 + 2 * 16
234 <x <298
That is, the interval would be (234, 298)
Answer:
78
Step-by-step explanation:
(-5)•(-5) + 3•21= 25+ 63= 78
We know, Volume of a Cylinder = πr² h/3
v = 3.14 * 4² * 6/3
v = 3.14 * 16 * 2
v = 100.48
In short, Your Answer would be Option B
Hope this helps!
Given:
In triangle KLM, KL = 123 cm and measure of angle K is 35 degrees.
To find:
The length of the side KM to the nearest tenth of a centimeter.
Solution:
In a right angle triangle,

In the given right triangle KLM,



Multiply both sides by 123.



The measure of side KM is 100.8 cm.
Therefore, the correct option is (2).