Consider a monohydrid cross, which is an Aa by Aa cross, where the expected outcome of genotype frequencies for the offspring is 1 to 2 to 1 using a Punnett square. You have observed 300 offspring with 80 AA, 165 Aa, and 55 aa genotypes. Please calculate the chi-squared statistic for a test of the null hypothesis that the observed genotype counts in the offspring are equal to their expectation of 1 to 2 to 1.

b

step-by-step explanation:

if a was wrong then it has to be b it cant be c either because not all the y intercepts are the same!

ok now , i would like brainlest plz

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.

yes the data was not prsented

step-by-step explanation:

  answer:

  write 548 as 500 + 40 + 8. multiply each addend by 5. then add the three products to get 2,740.