Total balance in Mr. Copley at the end of the month = $1473.61.
Initial deposite = $75 (not shown in his checkbook).
Let us assume his balance at the beginning of the month = x.
Total balance at the end of the month = The balance at the beginning of the month + total deposite.
$1473.61 = x + 75.
Subtracting 75 from both sides, we get
1473.61 -75 = x + 75 -75.
1398.61 = x.
Therefore, his balance at the beginning of the month was $1398.61.
Answer:
27
Step-by-step explanation:
33.33% = 1/3
1/3 x 81 = 27
If I'm right follow me on gram: Greyhasnoshame
As always, the sum of the measures of the angles in a triangle is 180°.
(x -5) +(3x +30) +35 = 180
4x +60 = 180 . . . . . . . . . . . . . collect terms
4x = 120 . . . . . . . . . . . . . . . . . subtract 60
x = 30 . . . . . . . . . . . . . . . . . divide by 4
a) The value of x is 30.
Answer:
1/16
Step-by-step explanation:
This question involves two distinct genes; one coding for seed shape and the other for cotyledon color. The alleles for round seeds (R) and yellow cotyledons (Y) are dominant over the alleles for wrinkled seed (r) and green cotyledon (y) respectively.
In a cross between a truebreeding (i.e. same alleles for both genes) pea having round seeds and yellow cotyledon (RRYY) and a truebreeding pea having wrinkled seeds and green cotyledon (rryy), the F1 offsprings will all possess a heterozygous round seed and yellow cotyledon (RrYy).
The F1 offsprings (RrYy) will produce the following gametes: RY, Ry, rY, and ry. Using these gametes in a punnet square (see attached image), 16 possible offsprings will be produced in a ratio 9:3:3:1.
According to the question, 3/16 of the F2 offsprings will possess round seeds and green cotyledons, however, only 1 of them will be truebreeding i.e. RRyy. Hence, 1/16 of the F2 offsprings will be truebreeding for round seeds and green cotyledons.