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gregori [183]
4 years ago
11

After attending a church cookout, a number of the attendees are admitted to the emergency room with nausea, vomiting, and diarrh

ea. After discovering all the individuals admitted to the ER consumed the egg salad at the event, the leftover egg salad was tested and found to be positive for salmonella. If 83 individuals consumed the egg salad and 49 individuals were admitted to the ER with salmonella, what was the cumulative incidence of salmonella assuming that the 49 individuals admitted to the ER were the only ones to be affected by the bacteria?
A. 0.410
B. 0.590

C. 0.059

D. 0.041
Mathematics
1 answer:
Helen [10]4 years ago
7 0

Answer:

0.590

Step-by-step explanation:

Cumulative frequency = number of new cases during a particular period / number of individuals at risk = number of people down with salmonella / total number of people present =49 /83 = 0.590

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The​ quality-control manager at a compact fluorescent light bulb​ (CFL) factory needs to determine whether the mean life of a la
stira [4]
<h2>Answer with explanation:-</h2>

Let \mu be the population mean .

By observing the given information, we have :-

H_0:\mu=7450\\\\H_a:\mu\neq7450

Since the alternative hypotheses is two tailed so the test is a two-tailed test.

We assume that the  life of a large shipment of CFLs is normally distributed.

(a) Given : Sample size :  n=81 , since n>30 so we use z-test.

Sample mean : \overline{x}=7240

Standard deviation : \sigma=1350

Test statistic for population mean :-

z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

\Rightarrow\ z=\dfrac{7240-7450}{\dfrac{1350}{\sqrt{81}}}\\\\\Rightarrow\ z=-1.4

The critical value (two-tailed) corresponds to the given significance level :-

z_{\alpha/2}=z_{0.025}=1.96

Since the observed value of z (-1.4) is less than the critical value (1.96) , so we do not reject the null hypothesis.

Hence, we conclude that we have enough evidence to accept that the mean life is different from 7450 hours .

(b) The p-value : 2P(z>-1.4)=0.1615 , it means that the probability that the life of CFLs less than 7240 and greater than 7240 is 0.1615.

(c) The confidence interval for population mean is given by :-

\overline{x} \pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=7240\pm(1.96)\dfrac{1350}{\sqrt{81}}\\\\=7240\pm294\\\\=(6946,7534)

,

8 0
3 years ago
Like b and c are parallel what is the measure of &lt;2
Vadim26 [7]
C
Hope this helps 8)
5 0
3 years ago
What are the coordinates of the hole in the graph of the function?
MakcuM [25]

Answer:

The hole is at (5,7)

Step-by-step explanation:

x^2 − 3x − 10

-------------------

x−5

Factor the numerator

(x-5)(x+2)

-------------------

x−5

There is a hole at x=5 since it will cancel in the numerator and the denominator

f(x) = x+2  and letting x = 5

f(5) = 5+2

The hole is at (5,7)

4 0
3 years ago
Find what percent of 570 is 190​
natali 33 [55]

Answer:

190/570

Step-by-step explanation:

190/570

8 0
3 years ago
4)
liberstina [14]
Arc length is the angle/360 times circumference. The diameter of the unit circle is 2. So the circumference of the unit circle is 2pi, if you use 3.14 for pi, then the circumference is 6.28. So your equation is
x/360 times 6.28=4.2, divide by 6.28, then multiply by 360
x=240.76432
Your answer rounded to the nearest thousanth is 240.764

6 0
3 years ago
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