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d1i1m1o1n [39]
3 years ago
12

What would it mean if there was an outbreak of cholera? A. The disease has infected a few people spread out over a large distanc

e. B. The disease has spread suddenly and rapidly. C. The disease has been eradicated or erased; it no longer exists. D. The disease has been controlled and is no longer a threat to humans. Please select the best answer from the choices provided A B C D
Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
5 0

Answer:

I think that it is B because its the only answer that makes sense

Step-by-step explanation:

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what is the prime factorization of 55^5 x 65 x 9^15 and why? A. 3^15 * 5^6 *11^5*13 B. 3^30 *5^6 *11^5 *13 C.3^30 * 5^6 *11 * 13
leonid [27]

Answer:

B

Step-by-step explanation:

1. Lets focus on 55^5

55^5 = 11^5 * 5^5

2. now on 65

65 = 5 * 13

3. now on 9^15

9^15=(3^2)^15 = 3^30

4. combine all three parts

11^5 * 5^5 * 5 * 13 * 3^30 = 11^5*5^6*13*3^30

so our answer is B

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From a random sample of 58 businesses, it is found that the mean time the owner spends on administrative issues each week is 21.
zzz [600]

Answer: (20.86, 22.52)

Step-by-step explanation:

Formula to find the confidence interval for population mean :-

\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}

, where \overline{x} = sample mean.

z*= critical z-value

n= sample size.

\sigma = Population standard deviation.

By considering the given question , we have

\overline{x}= 21.69

\sigma=3.23

n= 58

Using z-table, the critical z-value for 95% confidence = z* = 1.96

Then, 95% confidence interval for the amount of time spent on administrative issues will be :

21.69\pm (1.96)\dfrac{3.23}{\sqrt{58}}

=21.69\pm (1.96)\dfrac{1.7}{7.61577}

=21.69\pm (1.96)(0.223221)

\approx21.69\pm0.83

=(21.69-0.83,\ 21.69+0.83)=(20.86,\ 22.52)

Hence, the 95% confidence interval for the amount of time spent on administrative issues = (20.86, 22.52)

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3 years ago
3. Solve the equation c2 = 80.
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C=40 bahshshahajehsh
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