<span>area = </span><span />
<span>Therefore, </span>
<span /><span></span>
<span /><span>Width = 4cm
</span>
If y varies directly as x, and y=8 when x=3, find y when x=15.
1 answer:
Answer:
y=40
Step-by-step explanation:
The formula for Direct Variation is y=kx or k=y/x. In this case I would use k=y/x. If you're y is 8 and x is 3, this means that K=8/3. Using this, we know that X=15. We have to find a Y value so that the fraction with an x of 15 simplified is 8/3. To do this you would write 8/3 and y/15. Now cross multiply to get 3y=120. Divide by 3 to get y=40. View my attachment for the work!
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<u>Annotation</u>
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer
Answer:
E (Y) = 3
Step-by-step explanation:
If a 4-sided die is being rolled repeatedly; and the odd-numbered rolls (1st 3rd,5th, etc.)
The probability of odd number roll will be, p(T) =
However, on your even-numbered rolls, you are victorious if you get a 3 or 4. Also, the probability of even number roll, p(U) =
In order to calculate: E (Y); We can say Y to be the number of times you roll.
We know that;
E (Y) = E ( Y|T ) p(T) + E ( Y|U ) p(U)
Let us calculate E ( Y|T ) and E ( Y|U )
Y|T ≅ geometric =
Y|U ≅ geometric =
also; x ≅ geometric (p)
∴ E (x) =
⇒ = 4 ; also
= 2
E (Y) = 4 × + 2 ×
= 2+1
E (Y) = 3