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3 years ago

Please help! Correct answer only!

Mathematics

1 answer:

bogdanovich [222]3 years ago

6 0

Answer:

<em>Expected Payoff ⇒ $ 1.50 ; Type in 1.50</em>

Step-by-step explanation:

Considering that 1 out of the 100 tickets will have a probability of winning a 150 dollar prize, take a proportionality into account;

$100 - Number of Tickets,\\1 - Number of Tickets You Can Enter,\\\\1 / 100 - Probability of Winning,\\$ 150 - Money Won,\\\\Proportionality - 1 / 100 = x / 150, where x - " Expected Payoff "\\\\1 / 100 = x / 150,\\100 * x = 150,\\\\Conclusion ; x = 1.5 dollars$

<em>Thus, Solution ; Expected Payoff ⇒ $ 1.50</em>

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What is 3.0029 in short word form?

vovikov84 [41]

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Kathy runs a 26.2- mile marathon at an average pace of 6.2 minutes per mile how long will it take her to finish

drek231 [11]

Around 2:45:34
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Read 2 more answers

Use the law of sines to find the value of w.

Lunna [17]

Option B: 4.0 cm is the value of w

Explanation:

It is given that $\angle W= 39$ , $\angle U=31$ , $VW=3.3$ and $UV=w$

Using the law of sines, we have,

$\frac{sin U}{3.3} =\frac{sin W}{w}$

Substituting the values, we have,

$\frac{\sin 31}{3.3}=\frac{\sin 39}{w}$

Cross multiplying, we get,

$(sin31) w=3.3( \sin 39)$

Simplifying, we have,

$0.515w=2.077$

Dividing both sides by 0.515, we have,

$w=4.0$

Thus, the value of w is 4.0 cm

Hence, Option B is the correct answer.

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3 years ago

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The force of gravity on Mars is different than on Earth. The function of the same situation on Mars would be represented by the

sweet-ann [11.9K]

Answer:

If thrown up with the same speed, the ball will go highest in Mars, and also it would take the ball longest to reach the maximum and as well to return to the ground.

Step-by-step explanation:

Keep in mind that the gravity on Mars; surface is less (about just 38%) of the acceleration of gravity on Earth's surface. Then when we use the kinematic formulas:

$v=v_0+a\,*\,t\\y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2$

the acceleration (which by the way is a negative number since acts opposite the initial velocity and displacement when we throw an object up on either planet.

Therefore, throwing the ball straight up makes the time for when the object stops going up and starts coming down (at the maximum height the object gets) the following:

$v=v_0+a\,*\,t\\0=v_0-g\,*\,t\\t=\frac{v_0}{t}$

When we use this to replace the 't" in the displacement formula, we et:

$y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2\\y-y_0=v_0\,(\frac{v_0}{g} )-\frac{g}{2} \,(\frac{v_0}{g} )^2\\y-y_0=\frac{1}{2} \frac{v_0^2}{g}$

This tells us that the smaller the value of "g", the highest the ball will go (g is in the denominator so a small value makes the quotient larger)

And we can also answer the question about time, since given the same initial velocity $v_0$ , the smaller the value of "g", the larger the value for the time to reach the maximum, and similarly to reach the ground when coming back down, since the acceleration is smaller (will take longer in Mars to cover the same distance)

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3 years ago

HELP ILL GIVE BRAINLIEST!!!!

Setler [38]

Its A i had the same one. Hope it helps.

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