A scientest is studying a new super bacteria that is threatening to cause an outbreak. On the first day, there were 5 organisms
1 answer:
Answer:
C. 125
Step-by-step explanation:
Given:
Bacteria on day 1 = 5
On day 2, each bacteria got split into 5.
So, number of bacteria on day 2 =
Now, again on day 3, each of the 25 bacteria got split into additional 5 bacteria.
So, number of of bacteria on day 3 =
Therefore, the population of bacteria on day 3 is 125.
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Answer:
We conclude that this is an unusually high number of faulty modems.
Step-by-step explanation:
We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.
The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.
Let p = <em><u>population proportion</u></em>.
So, Null Hypothesis, : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}
Alternate Hypothesis, : p > 0.013 {means that this is an unusually high number of faulty modems}
The test statistics that would be used here <u>One-sample z-test</u> for proportions;
T.S. = ~ N(0,1)
where, = sample proportion faulty modems=
= 0.027
n = sample of modems = 367
So, <u><em>the test statistics</em></u> =
= 2.367
The value of z-test statistics is 2.367.
Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>
Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.
Therefore, we conclude that this is an unusually high number of faulty modems.
