Answer:
possible. it's a scalene triangle, an angle with different lengths
Answer: see below
Step-by-step explanation:
If this is a 6 sided die then the only possible roll that isnt an odd, a 2, or a 4 is 6.
The probability of something happening is also the probability of something not happening. So in order to not roll an odd, a 2 or a 4 in two rolls, you have to roll a 6 twice.
P = 1 - (1/6)(1/6) = 1 - (1/36) = (35/36)
Answer:
Step-by-step explanation:
According to intermediate value theorem, if a function is continuous on an interval ![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
, and if 
is any number between 
and 
, then there exists a value, 
, where 
, such that 
In the given question,
Intermediate Value Theorem is used to show that there is a root of the given equation in the specified interval.
Here,
Put 
Put 
Answer: I don’t understand
Step-by-step explanation:
Next time actually ask something about college math
Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
