Answer:
None of the equations are true for p = 3.4
Step-by-step explanation:
In order for the inequality to be valid we need to apply the value for p and check wether or not it is true. We gonna do for each of the following:
A. 3p < 10.2
3*3.4 < 10.2
10.2 < 10.2
Not true, since the left side is equal to the right side and not less.
B. 13.6 < 3.9p
13.6 < 3.9*3.4
13.6 < 13.26
Not true since 13.6 is greater than 13.26
C. 5p > 17.1
5*3.4 > 17.1
17 > 17.1
Not true since 17 is less than 17.1
D. 8.5 > 2.5p
8.5 > 2.5*3.4
8.5 > 8.5
Not true since the left side is equal to the right side.
Answer:
y=7600(5^(t/22))
Step-by-step explanation:
This is going to be an exponential function as it grows rapidly.
This type of question can be solved using the formula y=a(r^x), where a is the inital amount, r the factor by which the amount increases and x is the unit of time after which the amount increases.
x=t/22
a=7600
r=5
∴y=7600(5^(t/22))
V = PI x r^2 x H
1348.79 = 3.14 x 5.5^2 x H
1348.79 = 94.985 x h
h = 1348.79 / 94.985
h = 14.2 cm ( round answer if needed)
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)
