Answer:
The tag price would be 275 dollars.
Step-by-step explanation:
10% of 250.00 is 25. But the tag price is without the discount and the 250.00 is with the discount. So, if you add 25 to 250 you get 275.
We will be using the formula: Speed = Wavelength * Frequency
1. <u>Given:</u> Wavelength = 5 m and Frequency = 2 Hz
Speed of the wave = Wavelength * Frequency
Speed of the wave = 5 * 2
Speed of the wave = 10 m/s
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2. <u>Given:</u> Speed of the wave = 100 m/s and Frequency= 1000 Hz
Wavelength = Speed / Frequency
Wavelength = 100 / 1000
Wavelength = 0.1 m
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3. <u>Given:</u> Speed of the Wave = 25 m/s Wavelength = 5 m
Frequency = Speed / Wavelength
Frequency = 25 / 5
Frequency = 5 Hz
Answer:
I will!!!!!!
Step-by-step explanation:
ok
(25, 160) and (50, 320)
(y2-y1)/(x2-x1)
(320-160)/(50-25)
160/25 = slope
Y = 6.4x + b
Plug in points and solve for b
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
