A sick cow loses an average of 15 3/5 pounds each month.Represent the total change in the weight of the cow after 6 months

The line segment joining the points ( 3,-4) and (1,2) is trisected at the points P and Q . If the coordinates of P and Q are (p

sdas [7]

Answer:

P = 7/3 and q = 0

Step-by-step explanation:

The given parameters are;

The endpoints of the segment are, (3, -4) and (1, 2)

The points that trisect the given points are P and Q with coordinates (p, -2) and (5/3, q) respectively

Therefore, we have;

The first point of the trisection cuts 1/3 of the length from one point and the second point of the trisection cuts 2/3 of the length from the same point

The coordinates of P or Q = (3 - (3 - 1)/3, -4 - (-4 - 2)/3) = (7/3 , -2)

Therefore, given that the y-coordinate value of the derived point coincides with the y-coordinate value of the point P, (p, -2) and there is only one point with x = -2 on the line, we have that the coordinate of the point P is (p, -2) = (7/3 , -2)

∴ P = 7/3

Similarly we have the second point of the trisection, Q, given as follows;

We are already given the x-coordinate value of the point Q as the 5/3 in (5/3, q)

Point Q = (5/3, q) = (3 - 2×(3 - 1)/3, -4 - 2×(-4 - 2)/3) = (5/3 , 0)

Point Q = (5/3, q) = (5/3 , 0)

∴ q = 0.

3 0

3 years ago

Consider three boxes with numbered balls in them. Box A con- tains six balls numbered 1, . . . , 6. Box B contains twelve balls

murzikaleks [220]

Answer:

a) 0.73684

b) 2/3

Step-by-step explanation:

part a)

$P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)}$

Using conditional probability as above:

(A,B,C)

Cases for numerator when:

P( A is 1 and exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1)

= $(\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) = 0.048611111$

Cases for denominator when:

P( Exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1) + P(not 1, 1 , 1)

$= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) + (\frac{5}{6}*\frac{1}{12}*\frac{1}{4})= 0.0659722222$

Hence,

$P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)} = \frac{0.048611111}{0.06597222} \\\\= 0.73684$