What is being requested, if I'm not mistaken, is the number of permutations for placing each of the 8 beads on the vertices of the cubes;
In this case, we have 8 different beads and 8 possible locations for each of them;
So the number of permutations is:
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
Answer:
a. 
b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349
Step-by-step explanation:
a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.
b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.
c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842
d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166
e. The z-score related to 6.4 kg is 
and the z-score related to 7 kg is 
, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194
f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349
Answer:
reflected
Step-by-step explanation:
ΔXYZ was dilated, then reflected, to create ΔQAG.
We know that the angles of a triangle sum to 180°. For ΔABC, this means we have:
(4x-10)+(5x+10)+(7x+20)=180
Combining like terms,
16x+20=180
Subtracting 20 from both sides:
16x=160
Dividing both sides by 16:
x=10
This means ∠A=4*10-10=40-10=30°; ∠B=5*10+10=50+10=60°; and ∠C=7*10+20=70+20=90.
For ΔA'B'C', we have
(2x+10)+(8x-20)+(10x-10)=180
Combining like terms,
20x-20=180
Adding 20 to both sides:
20x=200
Dividing both sides by 20:
x=10
This gives us ∠A'=2*10+10=20+10=30°; ∠B'=8*10-20=80-20=60°; and ∠C'=10*10-10=100-10=90°.
Since the angle are all congruent, ΔABC~ΔA'B'C' by AAA.
