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3 years ago

What is the point probability

Mathematics

1 answer:

Lady_Fox [76]3 years ago

5 0

29.4% of the time, the point will be on the shaded region of the circle.

Step-by-step explanation:

Step 1:

To find the probability of any event we divide the number of favorable outcomes by the total number of outcomes.

Here the favorable outcome is the point being on the shaded region and the total number of outcomes is the point being on any point on the circle.

Suppose the circle is divided into 360 portions i.e. the angle of a circle is 360°.

Step 2:

The number of favorable outcomes = 106 (106° of the 360°).

The total number of outcomes = 360 (the entire 360°).

The probability of the point being on the shaded region = $\frac{106}{360} = 0.29444.$

This is equal to 29.444%. Rounding this off to the nearest tenth, we get 29.4%.

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the length of a rectangle is 10m more than its width. if the perimeter of rectangle is 80m find the dimensions of the rectangle

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The answers are 15 and 25

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4 years ago

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and

inn [45]

Complete Question

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.1 and the sample standard deviation was 1.2. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.)

t=8.169

P-value= ?

Answer:

a) $P-value=0$

b) Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/ cellulose composite exceeds 48 MPa.

Step-by-step explanation:

From the question we are told that:

Sample size $n=10$

Mean $\=x= 51.3$

Standard deviation $\sigma=1.2$

Significance level is taken as $\alpha=0.05$

t test statistics

$t=8.169$

Therefore

$P-Value=P(t>8.169)$

Critical point

$t_{\alpha,df}$

$\alpha=0.05$

$df=10-1=>9$

Therefore

P-value from T distribution table

$P-value=0$

Conclusion

$P-value (0)< \alpha(0.05)$

We Reject the Null Hypothesis $H_0$

Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/ cellulose composite exceeds 48 MPa.

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3 years ago

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3 years ago

For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subinterva

sergij07 [2.7K]

Given

we are given a function

$f(x)=x^2+5$

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

$\Delta x=\frac{b-a}{n}$

and the endpoints are given by

$a+k.\Delta x,\text{ for }0\leq k\leq n$

For k=0 and k=n, we get

$\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}$

Each rectangle has width and height as

$\Delta x\text{ and }f(x_k)\text{ respectively.}$

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

$Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)$

Here

$f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}$ $\Delta x=\frac{5-0}{n}=\frac{5}{n}$ $x_k=0+k.\Delta x=\frac{5k}{n}$ $f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5$

Now Area=

$\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}$

So the required area is 66.6 sq units.

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1 year ago

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