consider these graphs: Eric thinks the equation y=(x-3)^2 will match the blue graph. Jenney thinks it will match the red graph.
1 answer:
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Answer:
0.0764
Step-by-step explanation:
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.
In this problem, we have that:
What proportion of boxes is underweight (i.e., weigh less than 32 oz)?
This is the pvalue of Z when X = 32. So
has a pvalue of 0.0764, which is the correct answr
Answer:
Step-by-step explanation:
<u><em>First time drawing:</em></u>
Total outcome is 8
Favorable outcome is 2
The probability that the first marble drawing in random is blue is
=
<u><em>Second time drawing (with no replacement): </em></u>
Total outcome is 7 (8 - 1 = 7)
Favorable outcome is 2
The probability that the second marble drawing in random is yellow is
<em>The probability that the first marble is blue and the second marble is yellow is</em> ×
=
=
Answer:
The number of bacteria at initial = 187
Step-by-step explanation:
Given that the population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t.
Integrating both side we get
㏑ N = k t + C ------- (1)
Now given that after 3 hours it is observed that 500 bacteria are present and after 10 hours 5000 bacteria are present.
⇒ ㏑ 500 = 3 k + C -------- (2)
⇒ ㏑ 5000 = 10 k + C ------ (3)
⇒ ㏑ 5000 - ㏑ 500 = 7 k
⇒ ㏑ = 7 k
⇒ ㏑ 10 = 7 k
⇒ k = 0.329
Put this value of k in equation (2),
⇒ ㏑ 500 = 3 × 0.329 + C
⇒ C = 5.23
Put this value of C in equation 1 we get,
⇒ ㏑ N = k t + 5.23
Initially when t = 0 , then
⇒ ㏑ N = 5.23
⇒ N = 187
Thus the number of bacteria at initial = 187