Answer:
1. 0.8 cm
2. 1.6 cm
Step-by-step explanation:
1.
The scale for 2nd map is 1 cm to 50 km, that means "1 cm on map" is "50 km in real life".
We already know distance from Cleveland to Cincinnati is 40 km, which is less than 50, so we know the distance on map would be less than 1 cm.
So we set up ratio and figure out (let x be distance on map from Cleveland to Cincinnati):
Hene, 0.8 centimeters would be the distance in 2nd map
2.
A scale of 1:50 means 1 cm equal 50 cm
So, 0.8m would be
0.8 * 100 = 80 cm
Hence, 80 cm would be represented by 80/50 on the map, that is:
That is 1.6 centimeters
An odd number of factors were negative because if 1 number was negative and 2 were positive, positive times positive is positive. The answer times a negative number would be negative.
(positive times negative is negative0
I’m not sure what you want it rounded to
Height= 16.9913...
volume of a cone is v=1/3(pi)r^2h
So you’d do
1441.26=1/3(pi)9^2•h
I’m guess you are looking for 751?
Answer:
- P(E) = 1/2
- P(F) = 11/32
- P(G) = 1/6
- P(EF) = 5/52
- P(FG) = 1/32
- P(EG) = 1/6
Step-by-step explanation:
For the sum to be even, both dice can be odd, or both even. The probability of a dice being odd is 1/2 and the same is for it to be even. Since the result of the dices are independent, we have that
P(E) = (1/2)² + (1/2)² = 1/2
Out of the 36 possible outcomes for the dice (assuming that you can distinguish between first and second dice), there are 11 cases in which one dice is a 6 (if you fix 1 dice as 6, there are 6 possibilities for the other, but you are counting double 6 twice, so you substract one and you get 6+6-1 = 11). Since all configurations for the dices have equal probability, we get that
P(F) = 11/32
The probability for the second dice to be equal to the first one is 1/6 (it has to match the same number the first dice got). Hence
P(G) = 1/6
for EF, you need one six and the other dice even. For each dice fixed as 6 we have 3 possibilities for the other. Removing the repeated double six this gives us 5 possibilities out of 32 total ones, thus
P(EF) = 5/32
If one dice is 6 and both dices are equal, then we have double six, as a result there is only one combination possible out of 32, therefore
P(FG) = 1/32
If both dices are equal, in particular the sum will be even, this means that G= EG, and as a consecuence
P(EG) = P(G) = 1/6
