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Leya [2.2K]
3 years ago
11

What is .0196 rounded to the thousandths

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
8 0

so, the thousandth place is (EXAMPLE THIS IS: 0.00<u>1</u>) in this case, the 9 is in the thousandths place, and because the number to the right is greater than five, the answer is: 0.020, because you would round <u>UP</u>.

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A model rocket is launched with an intitial upupward velocity of 202 ft/s. The rocket’s height h (in feet) after seconds is give
dsp73

Given that the height of a object thrown within a gravitational field is given as a quadratic equation in time, <em>t</em>, the time at which the object is at a specified height can be found by the quadratic formula

The values of, <em>t</em>, for which the height of the rocket is 82 feet are;

t = 12.21 seconds or t = 0.42 seconds

Question: <em>Parts of the question that appear missing but can be found online are (i) To find the values of the time, t, when the rocket's height is 82 feet and round answers to the nearest hundredth</em>

The known parameters of the rocket are;

The initial upward velocity of the rocket, v = 202 ft./s

The given function representing the height, <em>h</em>, (in feet) after <em>t</em> seconds is presented as follows;

H = 202·t - 16·t²

The unknown;

The time for which the height of the rocket is 82 feet

Method;

Substitute H = 82 feet in the height function equation and solve for <em>t</em> as follows;

When H = 82, we get;

82 = 202·t - 16·t²

Therefore;

16·t² - 202·t + 82 = 0

Dividing the above equation by <em>2</em> gives;

(16·t² - 202·t + 82)/2 = 0/2

8·t² - 101·t + 41 = 0

By using the quadratic formula, we get;

\mathbf{t = \dfrac{101 \pm \sqrt{(-101)^2 - 4 \times 8 \times 41} }{2 \times 8}  =  \dfrac{101 \pm \sqrt{8889} }{16}}

Therefore, the values of <em>t</em> given by rounding off to the nearest hundredth are;

t = 12.21 or t = 0.42

The values of the time, <em>t</em>, at which the height of the rocket is 82 feet are t = 12.21 seconds or t = 0.42 seconds

Learn more about equation models of height as a function of time here;

brainly.com/question/84352

7 0
3 years ago
Compare and order the following numbers<br> 52-55,37 2
marshall27 [118]
Greatest to least is
52,37,2,-55 if you want least to greatest just reverse it
52,37,2 are whole numbers and natural numbers
While -55 is a negative number
They all are rational number
4 0
3 years ago
The daily revenues of a cafe near the university are approximately normally distributed. The owner recently collected a random s
lbvjy [14]

Answer:

The sample size to obtain the desired margin of error is 160.

Step-by-step explanation:

The Margin of Error is given as

MOE=z_{crit}\times\dfrac{\sigma}{\sqrt{n}}

Rearranging this equation in terms of n gives

n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2

Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as

n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n

As n is given as 40 so the new sample size is given as

n_2=4n\\n_2=4*40\\n_2=160

So the sample size to obtain the desired margin of error is 160.

4 0
3 years ago
The time needed to complete a final examination in a particular college course is normally distributed with a mean of 77 minutes
Komok [63]

Answer:

a. The probability of completing the exam in one hour or less is 0.0783

b. The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. The number of students will be unable to complete the exam in the allotted time is 8

Step-by-step explanation:

a. According to the given we have the following:

The time for completing the final exam in a particular college is distributed normally with mean (μ) is 77 minutes and standard deviation (σ) is 12 minutes

Hence, For X = 60, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=60−77 /12

Z=−1.4167

Using the standard normal table, the probability P(Z≤−1.4167) is approximately 0.0783.

P(Z≤−1.4167)=0.0783

Therefore, The probability of completing the exam in one hour or less is 0.0783.

b. In this case For X = 75, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=75−77 /12

Z=−0.1667

Using the standard normal table, the probability P(Z≤−0.1667) is approximately 0.4338.

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is obtained as follows:

P(60<X<75)=P(Z≤−0.1667)−P(Z≤−1.4167)

=0.4338−0.0783

=0.3555

​

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. In order to compute  how many students you expect will be unable to complete the exam in the allotted time we have to first compute the Z−score of the critical value (X=90) as follows:

Z=  X−μ /σ

Z=90−77 /12

Z​=1.0833

UsING the standard normal table, the probability P(Z≤1.0833) is approximately 0.8599.

Therefore P(Z>1.0833)=1−P(Z≤1.0833)

=1−0.8599

=0.1401

​

Therefore, The number of students will be unable to complete the exam in the allotted time is= 60×0.1401=8.406

The number of students will be unable to complete the exam in the allotted time is 8

6 0
3 years ago
S/4-6.8=-9.8 what does s equal
Hatshy [7]
S/4-6.8=-9.8
S/4= -9.8+6.8
S/4= -3
S=-3*4
S= -12
4 0
3 years ago
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