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finlep [7]
2 years ago
10

Atasha jumped 2 1/5 meter in a long jump competition. Arya jumped 1 2/3 meter. Who jumped longer and how many meters?​

Mathematics
1 answer:
Lorico [155]2 years ago
6 0

The person who jumped longer is Atasha by 8/15 meters

<u><em /></u>

<u><em>Who </em></u><u><em>jumped  longer?</em></u><u><em> </em></u>

In order to determine who jumped longer, compare the meters that was jumped by each person. Atasha jumped 2 1/5 meter and Arya jumped 1 2/3 meter. Looking at these fractions, the meters jumped by Atasha is greater than the meters jumped by Arya. This means that Atasha jumped longer.

<u><em>Difference in meters jumped </em></u>

In order to determine the difference, subtract the meters jumped by Arya from the meters jumped by Atasha.

2\frac{1}{5} - 1\frac{2}{3}

1\frac{3 - 10}{15}

\frac{18 - 10}{15}

\frac{8}{15}

To learn more about fractions, please check:

brainly.com/question/11664473?referrer=searchResults

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(-6x^4 + 53x^3 - 41x^2+ 6x + 16) + (x – 8) ​=  6x^{4} + 53x ^3 - 41x^2 +7x +8

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Brian "bicycle kicked" a soccer ball up from a height of 4ft off the ground with an initial upward velocity of 34 feet/sec. As t
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Answer:

1.95 seconds or 2.06 seconds

(Depending on your acceleration value)

Step-by-step explanation:

s = ut + 0.5at²

u = 34 ft/s

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s = 8-4 = 4 ft

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4 = 34t - 16.4042t²

16.4042t² - 34t + 4 = 0

t = 1.946, 0.125

Mike catches the ball the second time it was at a height of 8ft.

Do t = 1.95 s

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t = 2.05655765 (2.06 seconds in 3 sf)

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Ivahew [28]

Answer:

a)P(X=0)=(10C0) (0.15)^0 (1-0.15)^{10-0}= 0.1969

b) P(X=3)=(10C3) (0.15)^3 (1-0.15)^{10-3}= 0.1298

c) P(X \geq 1) = 1- P(X

P(X=0)=(10C0) (0.15)^0 (1-0.15)^{10-0}= 0.1969

So then we have:

P(X \geq 1) = 1- P(X

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.15)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

For this case we want this probability:

P(X=0)=(10C0) (0.15)^0 (1-0.15)^{10-0}= 0.1969

Part b

For this case we want this probability:

P(X=3)

And using the probability mass function we got:

P(X=3)=(10C3) (0.15)^3 (1-0.15)^{10-3}= 0.1298

Part c

For this case we want this probability:

P(X \geq 1)

And we can use the complenet rule and we got:

P(X \geq 1) = 1- P(X

P(X=0)=(10C0) (0.15)^0 (1-0.15)^{10-0}= 0.1969

So then we have:

P(X \geq 1) = 1- P(X

3 0
3 years ago
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