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4 years ago

Solve the formula for v1

Mathematics

1 answer:

Oksanka [162]4 years ago

6 0

So firstly, multiply both sides by t: $ta=v_1-v_0$

Next, add both sides by v0, and your answer will be: $ta+v_0=v_1$

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I need help with this

vazorg [7]

Zx = 5g(2x - c) Use the Distributive Property
zx = 10gx -5cg Subtract 10gx from both side
zx - 10gx = -5cg Factor out the x
x (z - 10g) = -5cg Divide both sides by (z-10g)
X= -5cg/z-10g

Hope this helps and please mark me as brainlest and like:)

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3 years ago

Find the length of the midsegment of the trapezoid.

Triss [41]

Answer:

Step-by-step explanation:

The Trapezoid Midsegment segment theorem states that the length of the midsegment is half the sum of the length of its two parallel sides. Therefore:

MN = ½(AB + DC)

MN = ½(21 + 37)

MN = ½(58)

MN = 29

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3 years ago

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inna [77]

Answer:

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Step-by-step explanation:

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4 years ago

Read 2 more answers

I dont understand do I multiply four to get the area?

UNO [17]

Answer:

yes

Step-by-step explanation:

3×5×45=365m²

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3 years ago

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A coin is tossed 20 times. A person, who claims to have extrasensory perception, is asked to predict the outcome of each flip in

Korolek [52]

Answer:

5.77% probability of being correct 14 or more times by guessing

Step-by-step explanation:

For each time the coin is tossed, there are only two possible outcomes. Either the person predicts the correct outcome, or she does not. The probability of predicting the correct outcome in a toss is independent of other tosses. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A coin is tossed 20 times.

This means that $n = 20$

Fair coin:

Equally as likely to be heads or tails, so $p = \frac{1}{2} = 0.5$

What is the probability of being correct 14 or more times by guessing

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 14) = C_{20,14}.(0.5)^{14}.(0.5)^{6} = 0.0370$

$P(X = 15) = C_{20,15}.(0.5)^{15}.(0.5)^{5} = 0.0148$

$P(X = 16) = C_{20,16}.(0.5)^{16}.(0.5)^{4} = 0.0046$

$P(X = 17) = C_{20,17}.(0.5)^{17}.(0.5)^{3} = 0.0011$

$P(X = 18) = C_{20,18}.(0.5)^{18}.(0.5)^{2} = 0.0002$

$P(X = 19) = C_{20,19}.(0.5)^{19}.(0.5)^{1} \approx 0$

$P(X = 20) = C_{20,20}.(0.5)^{20}.(0.5)^{0} \approx 0$

Then

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0370 + 0.0148 + 0.0046 + 0.0011 + 0.0002 + 0 + 0 = 0.0577$

5.77% probability of being correct 14 or more times by guessing

5 0

3 years ago