Pat needs to determine the height of a tree before cutting it down to be sure that it will not fall on a nearby fence. The angle of elevation of the tree from one position on a flat path from the tree is Upper H equals 60 degrees comma and from a second position Upper L equals 60 feet farther along this path it is Upper B equals 50 degrees . What is the height of the​ tree?

Answer:

The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Population:

Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000

Sample of 40

Shape approximately normal

Mean 350000

Standard deviation [tex]s = \frac{160000}{\sqrt{40}} = 25298[/tex]

The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.

Answer:

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

The mean would be:

[tex] \mu_{\bar X} =350000[/tex]

And the standard deviation would be:

[tex]\sigma_{\bar X} =\frac{160000}{\sqrt{40}}= 25298.221[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the selling price of a population, and for this case we know the following info

Where [tex]\mu=350000[/tex] and [tex]\sigma=160000[/tex]

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

The mean would be:

[tex] \mu_{\bar X} =350000[/tex]

And the standard deviation would be:

[tex]\sigma_{\bar X} =\frac{160000}{\sqrt{40}}= 25298.221[/tex]

Answer:

-17

Step-by-step explanation:

-2+3(8+-13)

-2+3(8-13)

-2+3(-5)

-2-15

-17

[tex]\text{Solve:}\\\\-2+3(8+-13)\\\\\text{Solve the variables inside of the parenthesis}\\\\-2+3(-5)\\\\-2-15\\\\\boxed{-17}[/tex]

Answer:

intercept=b0=75

Step-by-step explanation:

The least squares estimate of the intercept b0 can be computed as

b0=ybar-b1*xbar.

ybar=average number of shares of company stock=525.

xbar= average number of years employed=22.5.

slope=b1=20.

Thus,

intercept=b0=ybar-b1*xbar

intercept=b0=525-20*22.5

intercept=b0=525-450

intercept=b0=75.

Thus, the estimate of intercept b0=75.

Answer:

0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected

Step-by-step explanation:

Z-score

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

What is the probability that a randomly selected truck from the fleet will have to be inspected?

Values of Z above 2.5. So this probability is 1 subtracted by the pvalue of Z = 2.5.

Z = 2.5 has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected

Answer:

For this case we want this probability:

[tex] P(X > \mu +2.5 \sigma) [/tex]

And we can use the z score formula given by:

[tex] z = \frac{ X -\mu}{\sigma}[/tex]

And replacing we got:

[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]

We want this probability:

[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]

And using the normal standard table or excel we got:

[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the mileage of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]  

For this case we want this probability:

[tex] P(X > \mu +2.5 \sigma) [/tex]

And we can use the z score formula given by:

[tex] z = \frac{ X -\mu}{\sigma}[/tex]

And replacing we got:

[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]

We want this probability:

[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]

And using the normal standard table or excel we got:

[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]

Answer:

[tex]12 {x}^{2} + 15xy \\ = 3x(4x +5 y)[/tex]

hope this helps you...

Answer:

(a) [tex]X\sim N(0.10,\ 0.02^{2})[/tex].

(b) The probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.

(c) The 60th percentile is 11%.

Step-by-step explanation:

The complete question is:

The average THC content of marijuana sold on the street is 10%. Suppose the THC content is normally distributed with standard deviation of 2%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to two decimal places and give THC content in units of percent. For example, for a THC content of 11%, write "11" not 0.11.  

A. X ~ N( , )  

B. Find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11%.

C.Find the 60th percentile for this distribution.

Solution:

The random variable X is defined as the THC content for a bag of marijuana that is sold on the street.

(a)

The random variable X is Normally distributed with mean, μ = 0.10 and standard deviation, σ = 0.02.

Thus, [tex]X\sim N(0.10,\ 0.02^{2})[/tex].

(b)

Compute the value of P (X > 0.11) as follows:

[tex]P(X>0.11)=P(\frac{X-\mu}{\sigma}>\frac{0.11-0.10}{0.02})\\=P(Z>0.50)\\=1-P(Z<0.50)\\=1-0.69146\\=0.30854\\\approx 0.31[/tex]

*Use a z-table.

Thus, the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.

(c)

The pth percentile is a data value such that at least p% of the data-set is less than or equal to this data value and at least (100 - p)% of the data-set are more than or equal to this data value.

Let x represent the 60th percentile value.

The, P (X < x) = 0.60.

⇒ P (Z < z) = 0.60

The value of z for this probability is,

z = 0.26.

Compute the value of x as follows:

[tex]z=\farc{x-\mu}{\sigma}\\0.26=\farc{x-0.10}{0.02}\\x=0.10+(0.26\times 0.02)\\x=0.1052\\x\approx 0.11[/tex]

Thus, the 60th percentile is 11%.

Answer:

Using either method, we obtain:  [tex]t^\frac{3}{8}[/tex]

Step-by-step explanation:

a) By evaluating the integral:

 [tex]\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du[/tex]

The integral itself can be evaluated by writing the root and exponent of the variable u as:   [tex]\sqrt[8]{u^3} =u^{\frac{3}{8}[/tex]

Then, an antiderivative of this is: [tex]\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}[/tex]

which evaluated between the limits of integration gives:

[tex]\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}[/tex]

and now the derivative of this expression with respect to "t" is:

[tex]\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}[/tex]

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

[tex]g(x)=\int\limits^x_a {f(t)} \, dt[/tex]

is continuous on [a,b], differentiable on (a,b) and  [tex]g'(x)=f(x)[/tex]

Since this this function [tex]u^{\frac{3}{8}[/tex] is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

[tex]\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}[/tex]

Answer:

The process does not meets the specifications.

Step-by-step explanation:

It is provided that the Crunchy Potato Chip Company packages potato chips in a process designed for 10.0 ounces of chips.

The value of upper specification limit is, USL =  10.5 ounces.

The value of lower specification limit is, LSL = 9.5 ounces.

The mean weight of the chips bags is, [tex]\bar X=9.8\ ounces[/tex].

And the standard deviation  weight of the chips bags is, [tex]\sigma=0.12\ ounces[/tex].

Compute the value of process capability as follows:

[tex]C_{p_{k}}=min \{C_{p_{L}},\ C_{p_{U}}\}[/tex]

Compute the value of [tex]C_{p_{L}}[/tex] as follows:

[tex]C_{p_{L}}=\frac{\bar X-LSL}{3 \sigma}=\frac{9.8-9.5}{3\times 0.12}=0.833[/tex]

Compute the value of [tex]C_{p_{K}}[/tex] as follows:

[tex]C_{p_{K}}=\frac{USL-\bar X}{3 \sigma}=\frac{10.5-9.8}{3\times 0.12}=1.944[/tex]

The value of process capability is:

[tex]C_{p_{k}}=min \{C_{p_{L}},\ C_{p_{U}}\}[/tex]

      [tex]=min\{0.833,\ 1.944\}\\=0.833[/tex]

The value of process capability lies in the interval 0 to 1. So the  process is not capable of meeting design specifications.

Thus, the process does not meets the specifications.

Final answer:

The process capability of The Crunchy Potato Chip Company's packaging process has been questioned. By calculating the process capability index (Cpk), we find that the process capability is below the desirable threshold, suggesting that the process might not consistently meet the specified design requirements.

Explanation:

The student's query pertains to the process capability of The Crunchy Potato Chip Company's packaging process. This concept is part of quality control in process management and is measured by the process capability index (Cpk). In this case, the company has set specifications for a bag to contain between 9.5 and 10.5 ounces of chips and aims to have an average of 10.0 ounces. However, with a mean of 9.8 ounces and a standard deviation of 0.12 ounces, we need to calculate whether the process falls within the acceptable limits and how often it meets design specifications. To do this, the Cpk value is computed, which considers the mean, standard deviation, and the proximity of the mean to the closest specification limit.

Calculating the Process Capability Index (Cpk)

The formula for Cpk is:

Cpk = minimum of [(USL - mean) / (3 * sigma), (mean - LSL) / (3 * sigma)]

Where:

USL = Upper Specification Limit

LSL = Lower Specification Limit

mean = average net weight of the bags

sigma = standard deviation

Here, USL is 10.5 ounces, LSL is 9.5 ounces, mean is 9.8 ounces, and sigma is 0.12 ounces. Substituting these values:

Cpk = minimum of [(10.5 - 9.8) / (3 * 0.12), (9.8 - 9.5) / (3 * 0.12)]
Cpk = minimum of [1.9444, 0.8333]

The smaller of the two results is 0.8333, indicating that this is the Cpk value. A Cpk of 1.33 or higher is generally considered good, while a Cpk less than 1 indicates that the process may not be capable of meeting design specifications consistently. In this case, with a Cpk of 0.8333, The Crunchy Potato Chip Company's process may need improvement.

Answer:

She is wrong

Rectangle can be a trapezoid

Step-by-step explanation:

A trapezoid is a quadrilateral which has atleast one pair of parallel sides.

A rectangle has two pairs of parallel sides, so it is a trapezoid too.

All rectangles are trapezoid.

But all trapezoids are not rectangles

Answer:

the answers are

1) 1/16

2) -18

Answer:

[tex]1)32x = 2 \\ \frac{32x}{32} = \frac{2}{32} \\ x = \frac{1}{16} [/tex]

[tex]2)9 - \frac{1}{2} \\ \frac{9}{1} \frac{ \times }{ \times } \frac{2}{2} - \frac{1}{2} \\ \frac{18}{2} - \frac{1}{2} \\ = \frac{17}{2} \\ = 8 \frac{1}{2} [/tex]

Answer:

The smallest sample size needed is 49.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income.

This is n for which [tex]M = 500, \sigma = 2119[/tex]

So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]500 = 1.645*\frac{2119}{\sqrt{n}}[/tex]

[tex]500\sqrt{n} = 1.645*2119[/tex]

[tex]\sqrt{n} = \frac{1.645*2119}{500}[/tex]

[tex](\sqrt{n})^{2} = (\frac{1.645*2119}{500})^{2}[/tex]

[tex]n = 48.6[/tex]

Rounding up

The smallest sample size needed is 49.

Answer:

[tex]n=(\frac{1.640(2119)}{500})^2 =48.31 \approx 49[/tex]

So the answer for this case would be n=49 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=37500[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=2119[/tex] represent the sample standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =500 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.640[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.640(2119)}{500})^2 =48.31 \approx 49[/tex]

So the answer for this case would be n=49 rounded up to the nearest integer

Answer:

Probability of an Orange on the next toss

= (23/60) = 0.3833

Step-by-step explanation:

Orange: 46

Brown: 23

Green: 32

Yellow: 19

Probability of an Orange on the next toss

= n(orange colours obtained in the tosses) ÷ n(number of tosses)

n(orange colours in the tosses) = 46

Total number of tosses = 46 + 23 + 32 + 19

= 120.

Probability of an Orange on the next toss

= (46/120) = (23/60) = 0.3833

Hope this Helps!!!!

Answer:

3250 rubles.  

Step-by-step explanation:

1 USD = 32.5 rubles so...

1 * 100 USD = 32.5 * 100 rubles.  

The answer is 3250 rubles.  

Answer:

3250 rubles

Step-by-step explanation:

1 USD = 32.5 rubles

100 USD = 100 × 32.5

= 3250 rubles

Answer:

We can say that we are 90% confident that the population mean is between 29.876 and 170.124.

The upper number is 170.124.

Step-by-step explanation:

We have the sample's standard deviation, so we use the students' t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95([tex]t_{95}[/tex]). So we have T = 1.7531

The margin of error is:

M = T*s = 40*1.7531 =70.124

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 100 - 70.124 = 29.876

The upper end of the interval is the sample mean added to M. So it is 100 + 70.124 = 170.124

We can say that we are 90% confident that the population mean is between 29.876 and 170.124.

The upper number is 170.124.

Answer:

[tex]100-1.753\frac{40}{\sqrt{16}}=82.470[/tex]    

[tex]100+ 1.753\frac{40}{\sqrt{16}}=117.530[/tex]    

So on this case the 90% confidence interval would be given by (82.470;117.530)

And the upper value would be 117.530 for this case

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=100[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=40 represent the sample standard deviation

n=16 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

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

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=16-1=15[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,15)".And we see that [tex]t_{\alpha/2}=1.753[/tex]

Now we have everything in order to replace into formula (1):

[tex]100-1.753\frac{40}{\sqrt{16}}=82.470[/tex]    

[tex]100+ 1.753\frac{40}{\sqrt{16}}=117.530[/tex]    

So on this case the 90% confidence interval would be given by (82.470;117.530)    

Answer:

paired t-test

Step-by-step explanation:

Paired t-test is used to measure before and after effect. For example, a researcher wants to check the effect of new teaching method and he wants to compares marks of students before new teaching method and marks of students before new teaching method.

In the given example there are two groups participants before group activity and participants after group activity. A researcher wants to investigate that whether recently widowed spouses leave house after performing group activity or not. The observations in these two groups are paired naturally and we take the differences of these two groups are considered as random sample. This type of t-test is known as paired t-test.

Answer:

x  = 16

Step-by-step explanation:

The sum of measures of these three angles of any triangle is equal to 180 °

54 + 90 + (2x + 4) = 180

144 + 2x + 4 = 180

148 + 2x = 180

2x = 180 - 148

2x = 32

[tex]x= \frac{32}{2} \\\\x=16[/tex]

Focus on triangle KLM. Since JKLM is a rectangle, angle KLM is right. Since the sum of the interior angles of every triangle is 180, we deduce that

[tex]\hat{K}+\hat{L}+\hat{M}=180[/tex]

Now plug the values for each angle and you have

[tex]54+90+(2x+4)=180[/tex]

Move 54 and 90 to the right side to get

[tex]2x+4=36 \iff 2x=32 \iff x=16[/tex]

Answer:

a) 88.54% probability of a diameter between 3.8 in and 4.3 in

b) 25.14% probability of a diameter smaller than 3.9in

c) 90.82% probability of a diameter larger than 4.2 in

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 4, \sigma = 0.15[/tex]

(a) a diameter between 3.8 in and 4.3 in,

This is the pvalue of Z when X = 4.3 subtracted by the pvalue of Z when X = 3.8.

X = 4.3

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{4.3 - 4}{0.15}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772

X = 3.8

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{3.8 - 4}{0.15}[/tex]

[tex]Z = -1.33[/tex]

[tex]Z = -1.33[/tex] has a pvalue of 0.0918

0.9772 - 0.0918 = 0.8854

88.54% probability of a diameter between 3.8 in and 4.3 in

(b) a diameter smaller than 3 9 in,

This is the pvalue of Z when X = 3.9. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{3.9 - 4}{0.15}[/tex]

[tex]Z = -0.67[/tex]

[tex]Z = -0.67[/tex] has a pvalue of 0.2514

25.14% probability of a diameter smaller than 3.9in

(c) a diameter larger than 4.2 in

This is 1 subtracted by the pvalue of Z when X = 4.2. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{4.2 - 4}{0.15}[/tex]

[tex]Z = 1.33[/tex]

[tex]Z = 1.33[/tex] has a pvalue of 0.9082

90.82% probability of a diameter larger than 4.2 in

Let [tex]f(x,y,z)=x+y+z-8e^{xyz}[/tex]. The tangent plane to the surface at (0, 0, 8) is

[tex]\nabla f(0,0,8)\cdot(x,y,z-8)=0[/tex]

The gradient is

[tex]\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)[/tex]

so the tangent plane's equation is

[tex](1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8[/tex]

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by [tex]t[/tex], then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

[tex](1,1,1)t+(0,0,8)=(t,t,t+8)[/tex]

or [tex]x(t)=t[/tex], [tex]y(t)=t[/tex], and [tex]z(t)=t+8[/tex].

(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)

Final answer:

This detailed answer explains how to find the equation of the tangent plane and the normal line to a given surface at a specified point. The parametric equations of the normal line are [tex]\( (x(t), y(t), z(t)) = (t, t, 8 + t) \).[/tex]

Explanation:

To find the equations of the tangent plane and the normal line to the given surface x + y + z = 8exyz at the specified point 0, 0, 8) we first need to find the partial derivatives of x + y + z - 8exyz with respect to x, y, and z.

Given the surface equation:

[tex]\[ x + y + z = 8exyz \][/tex]

Let's denote the function as [tex]\(F(x, y, z) = x + y + z - 8exyz\).[/tex]

(a) To find the equation of the tangent plane, we use the gradient vector of F at the point 0, 0, 8 as the normal vector to the tangent plane.

The gradient vector of F is given by:

[tex]\[ \nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \][/tex]

Let's calculate the partial derivatives:

[tex]\[ \frac{\partial F}{\partial x} = 1 - 8eyz \][/tex]

[tex]\[ \frac{\partial F}{\partial y} = 1 - 8exz \][/tex]

[tex]\[ \frac{\partial F}{\partial z} = 1 - 8exy \][/tex]

Now, evaluate these partial derivatives at the point (0, 0, 8):

[tex]\[ \frac{\partial F}{\partial x} = 1 - 8e(0)(0) = 1 \][/tex]

[tex]\[ \frac{\partial F}{\partial y} = 1 - 8e(0)(0) = 1 \][/tex]

[tex]\[ \frac{\partial F}{\partial z} = 1 - 8e(0)(0) = 1 \][/tex]

So, the gradient vector of [tex]\(F\) at \((0, 0, 8)\) is \( \nabla F = (1, 1, 1) \).[/tex]

Therefore, the equation of the tangent plane is:

[tex]\[ x + y + z = 8 \][/tex]

(b) To find the equation of the normal line, we use the same gradient vector [tex]\( \nabla F = (1, 1, 1) \)[/tex] as the direction vector for the line.

Given the point \((0, 0, 8)\) on the surface, the parametric equations of the normal line are:

[tex]\[ x(t) = 0 + t(1) = t \][/tex]

[tex]\[ y(t) = 0 + t(1) = t \][/tex]

[tex]\[ z(t) = 8 + t(1) = 8 + t \][/tex]

So, the parametric equations of the normal line are [tex]\( (x(t), y(t), z(t)) = (t, t, 8 + t) \).[/tex]

Answer:

a) P(Job offer) = 0.65

b) P(S|J) = 0.862

P(M|J) = 0.143

P(W|J) = 0.0179

c) P(S|J') = 0.400

P(M|J') = 0.343

P(W|J') = 0.257

Step-by-step explanation:

- Let the event that she gets a job be J

- Let the event that she does not get the job be J'

- Let the event that she receives a strong recommendation be S

- Let the event that she receives a moderate recommendation be M

- Let the event that she receives a weak recommendation be W.

Given in the question,

P(J|S) = 80% = 0.8

P(J|M) = 40% = 0.4

P(J|W) = 10% = 0.1

P(S) = 0.70

P(M) = 0.20

P(W) = 0.10

a) How certain is she that she will receive the new job offer?

P(J) = P(J n S) + P(J n M) + P(J n W) (since S, M and W are all of the possible outcomes that lead to a job)

But note that the conditional probability, P(A|B) is given mathematically as,

P(A|B) = P(A n B) ÷ P(B)

P(A n B) is then given as

P(A n B) = P(A|B) × P(B)

So,

P(J n S) = P(J|S) × P(S) = 0.80 × 0.70 = 0.56

P(J n M) = P(J|M) × P(M) = 0.40 × 0.20 = 0.08

P(J n W) = P(J|W) × P(W) = 0.10 × 0.10 = 0.01

P(J) = P(J n S) + P(J n M) + P(J n W)

P(J) = 0.56 + 0.08 + 0.01 = 0.65

b) Given that she does receive the offer, how likely should she feel that she received a strong recommendation?

This probability = P(S|J)

P(S|J) = P(J n S) ÷ P(J) = 0.56 ÷ 0.65 = 0.862

i. a moderate recommendation?

P(M|J) = P(J n M) ÷ P(J) = 0.08 ÷ 0.65 = 0.143

ii. a weak recommendation?

P(W|J) = P(J n W) ÷ P(J) = 0.01 ÷ 0.65 = 0.0179

c) Probability that she doesn't get job offer, given she got a strong recommendation = P(J'|S)

P(J'|S) = 1 - P(J|S) = 1 - 0.80 = 0.20

Probability that she doesn't get job offer, given she got a moderate recommendation = P(J'|M)

P(J'|M) = 1 - P(J|M) = 1 - 0.40 = 0.60

Probability that she doesn't get job offer, given she got a weak recommendation = P(J'|S)

P(J'|W) = 1 - P(J|W) = 1 - 0.10 = 0.90

Total probability that she doesn't get job offer

P(J') = P(J' n S) + P(J' n M) + P(J' n W)

P(J' n S) = P(J'|S) × P(S) = 0.20 × 0.70 = 0.14

P(J' n M) = P(J'|M) × P(M) = 0.60 × 0.20 = 0.12

P(J' n W) = P(J'|W) × P(W) = 0.90 × 0.10 = 0.09

Total probability that she doesn't get job offer

P(J') = P(J' n S) + P(J' n M) + P(J' n W)

= 0.14 + 0.12 + 0.09 = 0.35

Given that she does not receive the job offer, how likely should she feel that she received a strong recommendation?

This probability = P(S|J')

P(S|J') = P(J' n S) ÷ P(J') = 0.14 ÷ 0.35 = 0.400

i. a moderate recommendation?

P(M|J') = P(J' n M) ÷ P(J') = 0.12 ÷ 0.35 = 0.343

ii. a weak recommendation?

P(W|J') = P(J' n W) ÷ P(J') = 0.09 ÷ 0.35 = 0.257

Hope this Helps!!!

She has 62% certainty of getting the job. If she gets it, there are high chances (~90%) that the recommendation was strong. In case of rejection, she's slightly more likely to have received a strong recommendation than a moderate or weak one.

The subject of this problem involves the understanding of conditional probabilities and the principle of total probability. Let's denote the events as follows:

a. The total probability that she will receive the job offer is calculated as follows:

P(B)=P(A1)P(B|A1)+P(A2)P(B|A2)+P(A3)P(B|A3) = 0.7*0.8+0.2*0.4+0.1*0.1=0.62, so she is 62% certain that she will receive the job offer.

b. If she does receive the offer, the probabilities that she got a strong, moderate and weak recommendation are calculated using Bayes' Theorem:

c. If she doesn't get the job, the probabilities are calculated similarly:

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Answer:

P(yellow, then orange WITHOUT replacement) = 0.0158

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

This problem asks:

The probability of first getting a yellow marble, then an orange marble, without replacemtn.

Yellow marble first:

3 yellow marbles out of 3+6+1+8+2 = 20 marbles.

So P(a) = 3/20.

Then the orange marble:

Now the yellow marble is retired.

2 orange marbles out of 19.

So P(b) = 2/19

Probability:

P(yellow, then orange WITHOUT replacement) = P(a)P(b) = (3/20)*(2/19) = 0.0158

The answer is:

Both have the same number of data points

Both means are between 3 and 4

Westwood has less variability than Middleton

Answer:

The right anwers are : A B and E

Step-by-step explanation:

right on edg!

Answer:

95 percent confidence interval for μ is determined by

(60.688 ,  69.312)

a) be the same

Step-by-step explanation:

Step (i):-

Given data a random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph.

The sample size is n= 121

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

Step (ii) :-

Given  The 95 percent confidence interval for μ is determined as

(61.08, 68.92)

we are to reduce the sample size to 100 (other factors remain unchanged) so

given The sample size is n= 100

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

Step (iii):-

95 percent confidence interval for μ is determined by

[tex](x^{-} - 1.96\frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} } )[/tex]

[tex](65 - 1.96\frac{22}{\sqrt{100} } , 65 + 1.96 \frac{22}{\sqrt{100} } )[/tex]

(65 - 4.312 ,65 +4.312)

(60.688 ,  69.312) ≅(61 ,69)

This is similar to (61.08, 68.92) ≅(61 ,69)

Conclusion:-

it become same as (61.08, 68.92)

In the given scenario, the 95 percent confidence interval for μ would become wider if the sample size is reduced from 121 to 100. This is due to the increased uncertainty in estimation as a result of a smaller sample size.

In statistics, a confidence interval is a range in which a population parameter is estimated to lie. The calculation of a confidence interval involves several factors, one of which is the sample size. The width of the confidence interval decreases as the sample size increases, assuming that all other factors remain the same.

In this case, the sample size is proposed to decrease from 121 to 100. As a result, the confidence interval for μ would become wider. This is because as the sample size decreases, the standard error increases, which in turn widens the confidence interval, allowing for more uncertainty in estimations.

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Answer:

a) 68% falls within 7.5 and 13.5

b) 95% falls within 4.5 and 16.5

c) 99.7% falls within 1.5 and 19.5

Step-by-step explanation:

Given that:

mean (μ) = 10.5, Standard deviation (σ) = 3.

The Empirical Rule (also called the 68-95-99.7 Rule)  for a normal distribution states that all data falls within three standard deviations. That is 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

Therefore using the empirical formula:

68% falls within the first standard deviation (µ ± σ) = (10.5 ± 3) = (7.5, 13.5)

95% within the first two standard deviations (µ ± 2σ) = (10.5 ± 2(3)) = (10.5 ± 6) = (4.5, 16.5)

99.7% within the first three standard deviations (µ ± 3σ) = (10.5 ± 3(3)) = (10.5 ± 9) = (1.5, 19.5)

Answer:

: The percentage of newborn infants who are exclusively breastfed at the time of hospital discharge.

Data Source: Rhode Island Department of Health, Center for Health and Data Analysis, Newborn Developmental Risk Screening Program Database and Maternal and Child Health Database.

Footnotes: The year indicated is the mid-year of a five-year period of data.

Step-by-step explanation:

Cramér's V value of .22 in the study indicates a small to medium association between whether the mother ate breakfast and the gender of her baby.

In the study, Cramér's V measuress the strength of association between two nominal variables, in this case, the mother's breakfast-eating habits and the gender of her baby. The value of Cramer’s V ranges from 0 (indicating no association) to 1 (indicating a perfect association)). Given the value of Cramér's V at .22 for this study, there is a small to medium association between whether a mother ate breakfast and the gender of her baby. It does not mean that 22% of the variation in baby gender can be explained by the mother's breakfast-eating habits. Rather, it quantifies the extent of the association between these two factors.

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Answer:

Y=1/2x +5

Step-by-step explanation:

(0,5) (-10,0)

0-5/-10-0=1/2

(0,5) (x, y)

Y-5/x-10=1/2(crossmultiply)

2y-10=x

2y= x +10(divide all sides by 2)

Y=1/2x+5

Answer: 0.8

Step-by-step explanation:

There are 5 possible equally sized sections meaning there is a 0.2 or 20% of landing on each section. We can subtract the probability of getting a squirrel from the whole thing (1 as probability always equals 1) to determine the probability of not squirrel.

P(squirrel) = 0.2

1 - P(squirrel) = P(not a squirrel)

1 - 0.2 = 0.8

Answer: the anserw is 0.8

Step-by-step explanation:

Answer:

Correct option:

The 95% confidence interval is wider than the 90% confidence interval.

Step-by-step explanation:

The (1 - α)% confidence interval for the population proportion is:

[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

The width of this interval is:

[tex]W=UL-LL\\=2\times z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

The width of the interval is directly proportional to the critical value.

The critical value of a distribution is based on the confidence level.

Higher the confidence level, higher will be critical value.

The z-critical value for 95% and 90% confidence levels are:

[tex]90\%: z_{\alpha /2}=z_{0.05}=1.645\\95\%: z_{\alpha /2}=z_{0.025}=1.96[/tex]

*Use a z-table.

The critical value of z for 95% confidence level is higher than that of 90% confidence level.

So the width of the 95% confidence interval will be more than the 90% confidence interval.

Thus, the correct option is:

"The 95% confidence interval is wider than the 90% confidence interval."

Answer:

y= -5x-13

Step-by-step explanation:

start with the linear equation

y= mx+b

indentify what you already have.

x= -4 and y= 7 and m(slope)= -5

We are solving for b, so plug in what you have.

7= -5(-4)+b

Simplify.

7= 20+b

Subtract 20 on both sides.

-13= b

Rewrite your equation with the x and y values.

y= -5x-13