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.

answer: yes i agree that you should not put it on/p>

step-by-step explanation: the reason for this is that it wastes other peoples time and also make them fail as well and that is never good.

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!

1.3 million hoops i believe