Given the following observed and expected data (total of 1000), using chi-squared calculate a p-value that corresponds with this data: Phenotype Observed Expected
cinnabar, vestigial 384 390
roof 408 390
cinnabar, roof, vestigial 63 70
Wild type 72 70
vestigial 32 35
cinnabar, roof 34 35
roof, vestigial 4 5
cinnabar 3 5

a. 025> p> 0.1
b. 0.9 > p >0.75
c. 0.05 > p
d. 0.75 > p>0.5
e. 0.95 >p >0.9

The answer is ""

Step-by-step explanation:

Eight phenotypes were present.  

Df is provided also by a number of phenotypes -1 The degree of freedom

For p-value 0,9, Chi-square is 2.83;

The p-value of 0.75 is 4.5. Chi-square

Chi-sqaure value is observed at 2.965.

That means 0.90>p-value>0.75.

the ratio of white tiles to blue tiles is 48: 16. the ratio in it's simplest form is 3: 1.

step-by-step explanation:

ratio: take the first number and add a colon next to it, then add the second number after that. make sure the ratio order matches up with the written problem!

simplest form: we can simplify the ratio 48 : 16 by dividing both terms by the greatest common factor (gcf).

the gcf of 48 and 16 is 16.

divide both terms by 16:

48 ÷ 16 = 3

16 ÷ 16 = 1

reducing to simplest form:

48 : 16 = 3 : 1

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.