lauren graphed the relationship between how many days had passed since she planted a cactus and it’s height (in centimeters)
2 answers:
See the explanation
<h2>Explanation:</h2><h2 />Hello! Remember you have to write complete questions in order to get exact and good answers. However, I found a similar question on the internet which is attached below and want to explain it to you. Here we know that the graph is a line that passes through the points So the slope is:
<em>From the slope we know that the plant grows 1 cm every two days.</em>
<h2>Learn more:</h2>Parallel lines: brainly.com/question/12169569
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Answer:
76.10% probability that between 19 and 28 circuits in the sample are defective
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be 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.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
So
P(19 ≤ X ≤ 28)
Using continuity correction, this is , which is the pvalue of Z when X = 28.5 subtracted by the pvalue of Z when X = 18.5. So
X = 28.5
has a pvalue of 0.9279
X = 18.5
has a pvalue of 0.1669
0.9279 - 0.1669 = 0.7610
76.10% probability that between 19 and 28 circuits in the sample are defective