Answer: c) y = ln(5x) - 2
<u>Step-by-step explanation:</u>

Answer:
The answer is (B)
Step-by-step explanation:
Try out each of the situations. It can't be A, since the 0.9 is less than 1, so it has to be decreasing. This leaves only B and C. Since decreasing by 90 percent is the same as multiplying by 10 percent, the only possible answer left is B.
Hope this helps!
Answer:
7)m=-4
8)p=1
9)v=-12
10)n=-7
11)-13=x
12)b=-4
13)-108=b
14)n=4
Step-by-step explanation:
Answer:
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

The percentage of cockroaches weighing between 77 grams and 83 grams
This is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 83. So
X = 83



has a pvalue of 0.7734
X = 77



has a pvalue of 0.2266
0.7734 - 0.2266 = 0.5468
Rounded to the nearest whole number, 55%
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.
LCM between 10 and 19 is 190.
First we can find the prime factorization of 10:
2 * 5.
Prime factorization of 19:
19.
2 * 5 * 19 = 190