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Lelu [443]
2 years ago
8

The track team is trying to reduce their time for a relay race. First they reduce their time by 2.1 minutes. Then they are able

to reduce that time by 1/10. If their final time is 3.96 minutes, what was their beginning time?
What was the track teams initial relay time in minutes?
Mathematics
1 answer:
klemol [59]2 years ago
3 0
Answer: 15.7 minutes

Step-by-step explanation:

Let x be the time in the beginning (in minutes).

Given: The track team is trying to reduce their time for a relay race.

First they reduce their time by 2.1 minutes.

Then they are able to reduce that time by 10

If their final time is 3.96 minutes, then

x-t1-t2= 3.6
x= 3.6+ t1+ t2
x= 3.6+ 2.1+ 10
x= 15.7

Hence, their beginning time was 15.7 minutes.
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Which equivalent four-term polynomial can be created using the X method?
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C.x2 + 12x + 4x + 48

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A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were
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Answer:

a) ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

b) 0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

Step-by-step explanation:

1) Data given and notation  

n=900 represent the random sample taken    

X=372 represent the students were pursuing liberal arts degrees

\hat p=\frac{372}{900}=0.413 estimated proportion of students were pursuing liberal arts degrees

\alpha=0.01 represent the significance level

z would represent the statistic (variable of interest)    

p_v represent the p value (variable of interest)    

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

The margin of error is given by:

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If we replace we have:

ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

And replacing into the confidence interval formula we got:

0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

7 0
2 years ago
A stone is dropped into a barrel of water and sinks to the bottom. The ball is completely covered by water. By how much does the
Rom4ik [11]

Answer:

Answer: 5/12 or approx. 0.42

Step-by-step explanation:

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Meaning, the volume of the ball and the volume of the increase in water in the barrel will be equal.

Volume of a sphere:

V_s=\frac{4}{3} \pi r^3

The radius is half the diameter. The diameter of the sphere is 10, thus:

r = d/2 = 10/2 = 5

5^3 = 5*5*5 = 25*5 = 125

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The volume of a cylinder, which will be the shape of the increase in water, is the following:

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Since we know these 2 volumes are the same, we'll set them equal to each other and solve the equation:

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\frac{4}{3} * 125 * \pi = 400 * h * \pi

We can get rid of pi by dividing both sides by pi:

\frac{4}{3} * 125 = 400 * h

And now divide both sides by 400:

\frac{4}{3} * \frac{125}{400}= h

Just reversing it and putting h to the left:

h = \frac{4}{3} * \frac{125}{400} = \frac{4 * 125}{3 * 400}

I see that I can divide both the numerator and the denominator by 4:

h =\frac{125}{3 * 100}= \frac{125}{300}

I'll now divide both the numerator and the denominator by 25 to simplify the expression:

h=\frac{125}{300} =\frac{125 / 25}{300 / 25} = \frac{5}{12}

Answer: The water rises by 5/12 (five twelfths).

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Answer:

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