Answer:
yés it is linear because
two variables that gives a straight line when plotted on a graph.
if it was correct tell me good or it was mistake tell me truth
Step-by-step explanation:
hope it was correct
Answer:
<u>Given</u>:
- R(x) = 59x - 0.3x²
- C(x) = 3x + 14
<u>Find the following</u>:
- P(x) = R(x) - C(x) = 59x - 0.3x² - 3x - 14 = -0.3x² + 56x - 14
- R(80) = 59(80) - 0.3(80²) = 2800
- C(80) = 3(80) + 14 = 254
- P(80) = 2800 - 254 = 2546
I hope it helps you get it right
If you take the cost times people then decide by new set of people.$22x47=____÷74= 13.972972973
Answer:
68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
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:

What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
This is the pvalue of Z when X = 8.6 subtracted by the pvalue of Z when X = 6.4. So
X = 8.6



has a pvalue of 0.8413
X = 6.4



has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds