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from the given graph, It is 200.

The cross that will be short is iv.

TT : 2Tt : tt so it is 1 : 2 : 1

The part labeled i is the male inflorescence also known as tassel.

Natural selection: this suggests the survival of individuals whose adaptive variations are more suited to the environment and whose traits are transmitted to subsequent generations. Natural selection is the primary explanation for evolutionary change.

Natural selection: this suggests the survival of individuals whose adaptive variations are more suited to the environment and whose traits are transmitted to subsequent generations. Natural selection is the primary explanation for evolutionary change.

Arrange in ascending order 2, 5, 5, 5, 6, 6, 6, 6, 8, 8 Position of median = N/2 = 10/2 = 5th and 6th ∴Median = (6+6) / 2 = 12/2 = 6 Mode the item that repeats itself most = 6 ∴Product of modal and median number = 6 x 6 = 36

Number of grapes = 6 Number of banana = 11 Number of oranges = 13 Total = 30 P(Grape) = 6/30, P(Banana) = 11/30 P(Orange) = 13/30 ∴P(Either grape or banana) = 6/30 + 11/30 = 17/30

students who made the trip = 2 + 3 + 4 + 5 + 6 + 8 + 1 = 29

Number of itineraries = 7P4 = \(\frac{7!}{(7-4)!}\) =\(\frac{7!}{3!}\) =\(\frac{7 \times 6 \times 5 \times 4 \times 3!}{3!}\) = 840

K + 2K + 3K + 75 + 105 = 360 (at a point) 6K + 180 = 360 6K = 360 - 180 6K = 180 k = 180/6 = 30º

P(At least 11 yrs) = P(11yrs) + P(12yrs) = \(\frac{27}{40}\) + \(\frac{7}{40}\) = \(\frac{34}{40}\) = \(\frac{17}{20}\)

Mean \(= \bar{x} = \frac{60}{6} = 10\) mean deviation \(= \frac{\sum|x-\bar{x}|}{n} = \frac{28}{6}\) = 4.7

Total mark scored = 200 ∴200 = 15 + 4y - 4 + 5y + 54 + 28 + 8 200 = 9y + 101 200 - 101 = 9y 99 = 9y ∴y = 11

\(^{10}C_6 = \frac{10!}{(10-6)!6!}=\frac{10!}{4!6!}\\ =\frac{(10\times 9\times 8\times 7 \times 6!)}{4\times 3\times 2\times 1\times 6!}\\ =210\)