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dalvyx [7]
3 years ago
12

We have seen that drinking tea appears to offer a strong boost to the immune system. In a study extending the results,1 blood sa

mple was taken on 5 participants before and after one week of drinking about five cups of tea a day (the participants did not drink tea before the study started). The before and after blood samples were exposed to e.coli bacteria, and the production of interferon-gamma, a molecule that fights bacteria, viruses, and tumors, was measured. Mean production went from 155 pg/mL before tea drinking to 448 pg/mL after tea drinking. The mean difference for the 5 subjects is 293 pg/mL with a standard deviation in the differences of 242. The paper implies that the use of the t-distribution is appropriate.
(a) Use the information to test whether the mean production of interferon-gamma as a response to bacteria is significantly higher after drinking tea than before drinking tea. Use a 5 % significance level.
(b) Give the test statistic and the p-value
Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places.
(c) Is there evidence that the mean production of interferon-gamma is higher after drinking tea?
Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
5 0

Answer:

Hey i dont know the answer bc I’m too dumb

Step-by-step explanation:

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taurus [48]

Answer:

(x)^2+(5)^2=(9)^2

Step-by-step explanation:

(a)^2+(b)^2=(c)^2

(x)^2+(5)^2=(9)^2

x^2+25=81

-25 -25

x^2=56

x=28

p.s. this ^2 means squared

3 0
3 years ago
Which of the following is not a factor of x^3 – 3x^2 – 4x + 12?
Slav-nsk [51]

Answer:

c. (x + 3)

Step-by-step explanation:

using factor theorem

if x - 3 is a factor then p(a) = 0

p(a)= x^3 - 3x^2 - 4x + 12

a.(x-3)

p(3) = (3)^3 - 3(3)^2 - 4(3) + 12

= 27 - 27 - 12 + 12

= 0

therefore x-3 is a factor

b.(x + 2)

p(-2) = (-2)^3 - 3(-2)^2 - 4(-2) + 12

= -8 -12 + 8 + 12

,= 0

therefore x + 2 is a factor

c.(x + 3)

p(-3) = (-3)^3 - 3(-3)^2 - 4(-3) + 12

= -27 -27 + 12 + 12

= -30

therefore x + 3 is not a factor

d.(x-2)

p(2) = (2)^3 - 3(2)^2 - 4(2) + 12

= 8 -12 - 8 + 12

= 0

therefore x - 2 is a factor

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Do you have a picture of the graph?
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