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algol13
4 years ago
15

Approximately 10.3% of american high school students drop out of school before graduation. choose 10 students entering high scho

ol at random. find the probability that all 10 stay in school and graduate. round your answer to the nearest thousandth.
Mathematics
1 answer:
Hatshy [7]4 years ago
4 0
The probability that a binomial experiment of n trials and p, the probability of success, produces an outcome, x is given by:

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

Given that a<span>pproximately 10.3% of american high school students drop out of school before graduation, thus p = 0.103.

If we choose 10 students (i.e. n = 10)

The </span><span>probability that all 10 stay in school and graduate means the probability that none of the 10 students dropped outP(0)={ ^{10}C_0}(0.103)^0(1-0.103)^{10} \\  \\ 1\times1\times(0.897)^{10}=0.337 is given by


</span>
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Please answer worksheet 50 points!! And I’ll mark brainliest
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Answer:

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Step-by-step explanation:

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3 years ago
Researchers are interested in the effect of a certain nutrient on the growth rate of plant seeding. Use a hydroponics grow proce
Rashid [163]

Answer:

The 95% confidence interval for the mean would be given by (56.604;68.596)

The 99% confidence interval for the mean would be given by (53.196;72.004)

Step-by-step explanation:

1) Previous concepts

When we compute a confidence interval for the mean, we are interested on the parameter population mean, and we use the info from the sample to estimate this parameter.

2) Basic operations

The sample mean can be calculated with the following formula

\sum_{i=1}^n \frac{x_i}{n}

Using excel we can use this function to calculate the mean:

=AVERAGE(54.2,59.8,61.8,63.3,65.1,71.4)

The value obtained is \bar X=62.6

In order to find the sample deviation we can use this formula

s=\sqrt{\sum_{i=1}^n \frac{(x_i-\bar X)^2}{n-1}}

And using excel we can use this function to calculate the sample standard deviation:

=STDEV.S(54.2,59.8,61.8,63.3,65.1,71.4)

The value obtained is s=5.713

The sample size for this case is n=6, n<30 so then is better use the t distribution to calculate the margin of error. First we need to calculate the degrees of freedom, on this case df=n-1=6-1=5

The formula for the confidence interval would be given by:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

3) Part a

If we want a 95% or 0.95 of confidence, then the value for the signficance is \alpha=1-0.95=0.05, and \alpha/2=0.025, and 1-\frac{\alpha}{2}=0.975 so we can find the critical t value with the following formula in excel:

=T.INV(0.975,5)

And we got t_{\alpha/2}= 2.571

And we can replace into equation (1) and we got:

62.6 \pm 2.571\frac{5.713}{\sqrt{6}}

And using excel with the following formulas we got:

=62.6-2.571*(5.713/SQRT(6)) = 56.604

=62.6+2.571*(5.713/SQRT(6)) = 68.596

So the 95% confidence interval for the mean would be given by (56.604;68.596)

4) Part b

If we want a 99% or 0.99 of confidence, then the value for the signficance is \alpha=1-0.99=0.01, and \alpha/2=0.005, and 1-\frac{\alpha}{2}=0.995 so we can find the critical t value with the following formula in excel:

=T.INV(0.995,5)

And we got t_{\alpha/2}= 4.032

And we can replace into equation (1) and we got:

62.6 \pm 4.032\frac{5.713}{\sqrt{6}}

And using excel with the following formulas we got:

=62.6-4.032*(5.713/SQRT(6)) = 53.196

=62.6+4.032*(5.713/SQRT(6)) = 72.004

So the 99% confidence interval for the mean would be given by (53.196;72.004)

8 0
3 years ago
A desk is on sale for 34%  off. The sale price is $363  .What is the regular price? I tried this problem so many times but I kee
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The answer is the regular price is $486.42
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3 years ago
Read 2 more answers
Pls help with this!!
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A)
Let X be the food expenditure of a family.
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3 years ago
Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va
Burka [1]

Answer:

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)

with

\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}

for some c in the interval (-x, x)

In the particular case f

<em>f(x)=cos(x) </em>

<em> </em>

we have

\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)

therefore

\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0

and the polynomial approximation of T5(x) of cos(x) would be

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}

for some c in (-x,x). So

\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}

and we must find the values of x for which

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652

Working this inequality out, we find

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0
3 years ago
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