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3 years ago

Need help with these math questions

Mathematics

1 answer:

PolarNik [594]3 years ago

5 0

Answer:

31200 m

Step-by-step explanation:

120 times 60 equals 7200 and 200 times 120 equals 24000 and add the two up you get 31200

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Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

ozzi

$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$

by using the integration formula

we get,

$\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c$

now put the value of t=\sin\theta in the above equation

we get,

$\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c$

hence proved

7 0

3 years ago

The lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of

RideAnS [48]

Answer:

The range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].

Step-by-step explanation:

We are given that the lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of 17 days.

Let X = lengths of pregnancies in a small rural village

SO, X ~ Normal( $\mu=262,\sigma^{2} = 17^{2}$ )

Here, $\mu$ = population mean = 262 days

$\sigma$ = standard deviation = 17 days

Now, the 68-95-99.7 rule states that;

68% of the data values lies within one standard deviation points.

95% of the data values lies within two standard deviation points.

99.7% of the data values lies within three standard deviation points.

So, middle 68% of most pregnancies is represented through the range of within one standard deviation points, that is;

[ $\mu -\sigma$ , $\mu + \sigma$ ] = [262 - 17 , 262 + 17]

= [245 days , 279 days]

Hence, the range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].

5 0

3 years ago

Suppose that x is a Normally distributed random variable with an unknown mean μ and known standard deviation 6. If we take repea

nata0808 [166]

Answer:

1.2

Step-by-step explanation:

Given that X is Normally distributed random variable with an unknown mean μ and known standard deviation 6

Hence we can say for a sample of size 100, the sample mean will have a std deviation of = $\frac{6}{\sqrt{100} } =0.6$

Since population std deviation is known we can use Z critical value for finding out the confidence interval

For 95% using (68-95-99.7 rules) we have z critical value =2

Hence margin of error =2(std error) = 1.2

Confidence interval 95%

Lower bound = Mean - margin of error = Mean -1.2

UPper bound = Mean +1.2

Hence , 95% of all of these values of x should lie within a distance of __1.2___ from μ .

4 0

3 years ago

Someone please help me with this questions

insens350 [35]

Answer:

Coefficient Is the small 2

7 0

3 years ago

You want to learn about the Ferris–wheel riding habits of people, so you ask those leaving a theme park, “How many times did you

Korolek [52]

Answer:

Yes

Step-by-step explanation:

A statistical question is one that can be asked and expected to receive answers that may vary - in other words, where the data from each participant is different.

Here, the question is "How many times did you ride the Ferris wheel today?” Obviously, it's almost impossible that each person asked rode the Ferris wheel the same number of times. Some people may say "once", others may say "three times", and still others might not have ridden it at all.

Given the differences in all these answers, we can say that this IS a statistical question.

Hope this helps!

4 0

3 years ago

Read 2 more answers