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Strike441 [17]
3 years ago
11

To examine the potential effects on the transmission of dengue, Bian et al. (2010) infected mosquitoes from both the WB1 (labora

tory created) strain and the original wild strain with dengue. Fourteen days later, the mosquitoes were allowed to feed on an artificial food solution for 90 minute. Viral titers in the food solution were then measured in plague feeding units per ml. The results are given below. Mosquitoes were tested in groups, and so each data value is an average for the group, measured in plaque feeding units per ml. The data have already been log-transformed using log(Y+1), to better visualize the differences.
WB1: n1 = 8 R1 = 70.5
Wild: n2 = 8 R2 = 16

H0: the distribution of viral titers is the same between the two strains
HA: the distribution of viral titers is not equal

The data has been ranked from smallest to largest andthe rank sum (R1, R2) for each data set are listed above. Carry out a Mann-Whitney U-test on the WB1 data and solve for U1. Answer to one decimal place.
Mathematics
1 answer:
nekit [7.7K]3 years ago
6 0

Answer:

reject H0

the viral titer is lower in WB1

Step-by-step explanation:

yeah

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3 years ago
When 5655 is divided by a positive two digit intreger "N" the remainder is 11, when 5879 is divided by the same intreger N the r
Temka [501]

Answer:

8

Step-by-step explanation:

Hi,

When we divide 5655 by N, we get remainder of 11, which means that 5655-11 is a multiple of N.

5655 - 11 = 5644 is a multiple of N.

Similarly, 5879-14 should be a multiple of N.

5879 - 14 = 5865 is a multiple of N.

Because 5644 and 5865 are both multiples of N, their difference must be a multiple of N.

5865 − 5644 = 221 then 221 is a multiple of N.

We have three number of which N can be a multiple of, however we choose to factorize the smallest possible number amongst these three, which is 221. (This is only for simplification of the solution, smaller the number, less the factors)

221 : 1, 13, 17, 221.

There are only two two - digit factors: 13 and 17.

We divide 5865 and 5644 by both numbers.

\frac{5865}{13} = 451.15 \\\frac{5865}{17} = 345 \\\frac{5644}{13} = 434.15 \\\frac{5644}{17} = 332\\

Looking at these results, we know only 17 divides all three numbers.

Hence N=17.

The sum of both digits will be: 1 + 7 = 8

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