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love history [14]

3 years ago

15

A research scientist studied the effects of various treatments on the growth of poplar trees. He decided to treat a large sample

of saplings with a fertilizer, continual irrigation, irrigation and fertilizer and a control group (non-continual irrigation and no fertilization). He placed one quarter of all saplings into each of the groups and then measured their mass differential after 3 weeks. Did the treatment have an effect on the growth of the trees

1 answer:

STALIN [3.7K]3 years ago

4 0

Answer:

hello your question is incomplete attached below is the screenshot of the missing part

answer : The treatment had an effect on the growth of the trees

Explanation:

H0 : No significant effect

Ha : At least one treatment has a significant effect

To confirm the above hypothesis ( conduct one way ANOVA )

From the one way Anova

( X represents each group; from control to Fertilizer and irrigation )

∑ X1^2 = 0.2338 , ∑ X2^2 = 1.8284, ∑X3^2 = 0.1982, ∑X4^2 = 11.8492

Also : ∑ N = 5 + 5 + 5 + 5 = 20 where N = number of trees in each group

also K represents number of groups i.e. = 4

Next calculate the sum of squares ( calculated using online tools )

i) Between sum of squares: value = 4.6824

ii) Within sum of squares : value = 4.3572

Next step : Calculate the degrees ( DF)

i) Between degrees of freedom

Dfb = k - 1 where K = 4 therefore Dfb = 3

ii) within degrees of freedom

Dfw = N - k , where N = 20 and K = 4 therefore Dfw = 16

Next : calculate mean sum of squares

i) Between mean sum of squares

= 4.6824 / 3 = 1.5608

ii) within mean sum of squares

= 4.3572 / 16 = 0.2723

Calculate the F-statistic value

$F_{calculated }$ = Between mean sum of squares / within mean sum of squares

= 1.5608 / 0.2723 = 5.73

determine the F critical value

Fcritical = $F_{3,16,0.05}$ = 3.2389 ( obtained from F table )

since $F_{calculated } > F_{critical}$ hence we can say The treatment had an effect on the growth of the trees i.e. Reject H0

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If the frequency of the recessive allele for a gene is 0.3, calculate the expected frequency of heterozygotes in the next genera

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2 years ago

Calculate the final concentration of BSA in the problems below using the formula

Rudiy27

Answer:

A. $C_2=1.5\frac{mg}{mL}$

B. $C_2=0.075\frac{mg}{mL}$

C. $C_2=0.01\frac{mg}{mL}$

D. $C_2=0.001\frac{mg}{mL}$

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

$C_2=\frac{C_1V_1}{V_2}$

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

$C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}$

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

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C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}$

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}$

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