A production process fills containers by weight. Weights of containers are approximately normally distributed. Historically, the standard deviation of weights is 5.5 ounces. (This standard deviation is therefore known.) A quality control expert selects n containers at random. How large a sample would be required in order for the 99% confidence interval for to have a length of 2 ounces?
a. n  15
b. n  16
c. n  201
d. n  226

Rotate the right triangle with hypotenuse √7 about one leg to create a cone. Use the Pythagorean theorem to express the radius as a function of the other leg. To find the max volume, differentiate the volume formula and set equal to zero to solve for the radius and height.

The problem presents a right triangle with hypotenuse of √7 meters. This right triangle is rotated around one of its legs to form a right circular cone. To find the radius, height, and volume of the cone, we can use the properties of the triangle.  

Imagine that the triangle is rotated about the shorter leg. The height of the cone (h) is the length of the shorter leg, and the radius (r) is the longer leg. By the Pythagorean theorem, a² + b² = c², where c is the hypotenuse, a and b represent the legs of the triangle. Assume that a is the longer leg and b is the shorter leg.

From the Pythagorean theorem we get a² = (√7)² - b² = 7 - b² and so a = √(7 - b²).

The volume of the cone V = (1/3)πr²h = (1/3)π(√(7 - b²))²b = (1/3)π(7 - b²)b. Differentiating and setting the derivative equal to zero provides the desired maximum volume, giving us the values for the radius and height of the cone.

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Answer:

Is plausible that the successive throws are independent

Step-by-step explanation:

1) Table with info given

The observed values are given by the following table

__________________________________________________

First shot          Made          Second shot missed           Total

__________________________________________________

Made                  152                       33                                185

Missed                37                         8                                  45

__________________________________________________

Total                    189                       41                                230

2) Calculations and test

We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:

Null hypothesis: Independence between two successive free throws

Alternative hypothesis: No Independence between two successive free throws

_____________________________________________________

First shot                    Made                                    Second shot missed

_____________________________________________________

Made                  189(185)/230=152.0217                41(185)/230=32.9783

Missed                189(45)/230=36.9783                  41(45)/230=8.0217

_____________________________________________________

On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.

3) Calculate the chi square statistic

The statistic for this case is given by:

[tex]\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}[/tex]

Where O represent the observed values and E the expected values. Replacing the values that we got we have this

[tex]\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881[/tex]

Now with the calculated value we can find the degrees of freedom

[tex]df=(r-1)(c-1)=(2-1)(2-1)=1[/tex] on this case r means the number of rows and c the number of columns.

Now we can calculate the p value

[tex]p_v =P(\chi^2 >0.00008881)=0.9925[/tex]

On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.

Answer:

(-0.0861, 0.4023)

Step-by-step explanation:

We have large sample sizes [tex]n_{x} = 56[/tex] and [tex]n_{y} = 41[/tex]. A [tex]100(1-\alpha)[/tex]% confidence interval for the difference [tex]p_{x}-p_{y}[/tex] is given by [tex](\hat{p}_{x}-\hat{p}_{y})\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_{x}(1-\hat{p}_{x})}{n_{x}}+\frac{\hat{p}_{y}(1-\hat{p}_{y})}{n_{y}}}[/tex]. [tex]\hat{p}_{x}=43/56 = 0.7679[/tex] and [tex]\hat{p}_{y}=25/41=0.6098[/tex]. Because we want a 99% confidence interval for the difference [tex]p_{x}-p_{y}[/tex], we have that [tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.5758[/tex] (The area above 2.5758 and below the curve of the stardard normal density is 0.005) and the confidence interval is [tex](0.7679-0.6098)\pm (2.5758)\sqrt{\frac{0.7679(1-0.7679)}{56}+\frac{0.6098(1-0.6098)}{41}}[/tex] = (-0.0861, 0.4023).

Answer:

Part A)  Not collide

Part B)  k = 4

Part C)  Particle B is moving fast.

Step-by-step explanation:

Two particles move in the xy-plane. At time, t

Position of particle A:-

[tex]x(t)=5t-5[/tex]

[tex]y(t)=2t-k[/tex]

Position of particles B:-

[tex]x(t)=4t[/tex]

[tex]y(t)=t^2-2t+1[/tex]

Part A)  For k = -6

Position particle A, (5t-5,2t+6)

Position of particle B, [tex](4t,t^2-2t-1)[/tex]

If both collides then x and y coordinate must be same

Therefore,

5t - 5 = 4t    

       t = 5

[tex]2t+6=t^2-2t-1[/tex]

[tex]t^2-4t-7=0[/tex]

[tex]t=-1.3,5.3[/tex]

The value of t is not same. So, k = -6 A and B will not collide.

Part B) If both collides then x and y coordinate must be same

5t - 5 = 4t    

       t = 5

[tex]2t-k=t^2-2t-1[/tex]

Put t = 5

[tex]10-k=25-10-1[/tex]

[tex]k=4[/tex]

Hence, if k = 4 then A and B collide.

Part C)

Speed of particle A, [tex]\dfrac{dA}{dt}[/tex]

[tex]\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]

[tex]\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4[/tex]

Speed of particle B, [tex]\dfrac{dB}{dt}[/tex]

[tex]\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]

[tex]\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}[/tex]

At t = 5

[tex]\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2[/tex]

Hence, Particle B moves faster than particle A

To determine if the particles collide, we set up equations with their x-coordinates and y-coordinates. Part (a) asks if they collide when k = -6. Part (b) asks for the value of k that guarantees a collision, and part (c) compares the speeds at the time of collision.

To determine if the particles collide, we need to find out if their x-coordinates and y-coordinates are equal at any given time. Given the positions of particles A and B, we can set up two equations by equating their x-coordinates and y-coordinates. For part (a), when k = -6 we can solve the equations to find if they intersect. For part (b), we need to find the value of k that makes the two particles collide. Finally, for part (c), we compare the speeds of particles A and B when they collide.

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Answer:

x=-3+4t\\

y =5-7t

Step-by-step explanation:

we are given that when two point (x1,y1) and (x2,y2) are joined the parametric equation representing the line segment would be

[tex]x = x_1 + (x_2 - x_1)t y = y_1 + (y_2 - y_1)t, 0 \leq ≤t \leq ≤ 1[/tex]

Our points given are

[tex](-3, 5) to (1, -2)[/tex]

So substitute the values to get

[tex]x=-3+(1+3)t =-3+4t\\y = 5+(-2-3)t =5-7t[/tex]

Hence parametric equations for the line segment joining the points

(-3,5) and (1,-2) is

[tex]x=-3+4t\\y =5-7t[/tex]

Final answer:

The parametric equations for the line segment from (-3, 5) to (1, -2) are x = -3 + 4t and y = 5 - 7t for 0 ≤ t ≤ 1.

Explanation:

The question asks for parametric equations to describe the line segment joining the points P1(-3, 5) and P2(1, -2). According to the formula provided, we have:

x = x1 + (x2 – x1)t

y = y1 + (y2 – y1)t

Substituting the values of P1 and P2 into the equations:

x = -3 + (1 – (-3))t = -3 + 4t

y = 5 + (-2 – 5)t = 5 - 7t

This means for 0 ≤ t ≤ 1, the parametric equations representing the line segment from (-3, 5) to (1, -2) are:

x = -3 + 4t

y = 5 - 7t

Answer:20 hoseholds, what is the probability that more than

Step-by-step explanation:

Answer:

Step-by-step explanation:

To find median and mode for

a) In a uniform distribution median would be

(a+b)/2 and mode = any value

b) X is N

we know that in a normal bell shaped curve, mean = median = mode

Hence mode = median = [tex]\mu[/tex]

c) Exponential with parameter lambda

Median = [tex]\frac{ln2}{\lambda }[/tex]

Mode =0

The median of a distribution is the middle value while the mode is the highest occuring value

The median (M) of a uniform distribution is:

[tex]M = \frac{a +b}2[/tex]

A uniform distribution has no mode

For a normal distribution with the given parameters, we have:

Median = Mean = Mode = μ

Hence, the median and the mode are μ

For an exponential distribution with the given parameter, we have:

[tex]Median = \frac{\ln 2}{\lambda}[/tex]

The mode of an exponential distribution is 0

Read more about median and mode at:

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Answer:

;-; I belive it's the 1st or 2nd one

Step-by-step explanation:

hope this helps you.

ask me for explanation if you want and I'll try to explane it the best I Can.

(P.S;please mark me as brainlyest,ty in advaince)

Answer:

a)The sample size

Step-by-step explanation:

t - student distribution should be used whe we are facing a Normal Distribution population andwhen a sample size n is below 30.

t- student distribution is also a associated to a bell shape but is more flat and the values are more spread, tails are much more wide.

t- student use the concept of degree of fredom ( each degree of fredom correspont to an specific curve). As degree of fredom increase the curves become more close to a bell shape. For  n = 30 t-student curve is  a bell shape one

Answer:

To conduct the hypothesis test I would use the standard normal distribution to calculate the critical value and the p-value.

Step-by-step explanation:

To conduct the hypothesis test I would use the standard normal distribution because there is a large sample size of n = 125 households. This because a point estimator for the true proportion p of one-person households is [tex]\hat{p} = Y/n[/tex] which is normally distributed with mean p and standard error [tex]\sqrt{p(1-p)/n}[/tex] when the sample size n is large. Here Y is the random variable that represents the number of one-person households observed. Then the test statistic is [tex]Z = \frac{\hat{p}-0.27}{\sqrt{p(1-p)/n}}[/tex] which has a standard normal distribution under the null hypothesis.

Answer:

6x^2 + 18x + 19.

Step-by-step explanation:

5x^2 + 6x - 17 + (x + 6)^2

= 5x^2 + 6x - 17 + x^2 + 12x + 36

=  6x^2 + 18x + 19.

Answer:

6x^2 + 18x + 19

Step-by-step explanation:

The standard form of a polynomial depends on the degree of the polynomial. The polynomial is in the standard form when it is arranged such that the first term contains the highest degree, and it decreases with the consecutive terms.

The square of the binomial is (x+6)^2

(x+6)^2 = (x+6)(x+6) = x^2 + 6x + 6x + 36 = x^2 + 12x + 36

The sum of the two polynomials will be 5x^2+6x−17 + x^2 + 12x + 36

Collecting like terms,

5x^2 + x^2 + 6x + 12x + 36 -17

= 6x^2 + 18x + 19

Answer: 0.0026

Step-by-step explanation:

Let p denotes the proportion of students plan to go into general practice.

As per given , we have

Alternative hypothesis : [tex] H_a: p>0.28[/tex]

Since the alternative hypothesis [tex](H_a)[/tex] is right-tailed so the test is  a right-tailed test.

Also , it is given that ,

i.e. sample size : = 130

x= 490

[tex]\hat{p}=0.39[/tex]

Test statistic(z) for population proportion :

[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

, where p=population proportion.

[tex]\hat{p}[/tex]= sample proportion

n= sample size.

[tex]z=\dfrac{0.39-0.28}{\sqrt{\dfrac{0.28(1-0.28)}{130}}}\\\\=\dfrac{0.11}{0.0393798073988}=2.79330975101\approx2.79[/tex]

P-value for right-tailed test = P(z>2.79)=1-P(z≤ 2.79)  [∵P(Z>z)=1-P(Z≤z)]

=1- 0.9974=0.0026  [using z-value table]

Hence, the  P-Value for a test of the school's claim = 0.0026

To find the p-value for a test of the school's claim, we can use a hypothesis test. The null hypothesis (H0) is that the proportion of students planning to go into general practice is equal to or less than 28%. The alternative hypothesis (Ha) is that the proportion of students planning to go into general practice is greater than 28%. Given that 39% of the sample of 130 students plan to go into general practice, we calculated the test statistic and the p-value to determine that the p-value is 0.3078.

To find the p-value for a test of the school's claim, we can use a hypothesis test. The null hypothesis (H0) is that the proportion of students planning to go into general practice is equal to or less than 28%. The alternative hypothesis (Ha) is that the proportion of students planning to go into general practice is greater than 28%.

Given that 39% of the sample of 130 students plan to go into general practice, we can calculate the test statistic and the p-value. Using a chi-square test, we calculate the test statistic to be approximately 1.307. The degrees of freedom for this test is 1.

Looking up the critical value for a one-tailed test with an alpha level of 0.05 and 1 degree of freedom, the critical value is approximately 3.841. Since the test statistic is less than the critical value, we fail to reject the null hypothesis. Therefore, the p-value is greater than 0.05.

Therefore, the correct answer for the P-value is 0.3078.

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Answer:

[tex]\frac{7}{4}[/tex]×x

Step-by-step explanation:

given ratio of yellow to red flowers is 4:7.

let number of yellow flowers in Lindsay’s vase  is x

let number of red flowers in Lindsay’s vase  is y

[tex]\frac{x}{y}[/tex]=[tex]\frac{4}{7}[/tex]

number of  red flowers such that x is known is y= [tex]\frac{7}{4}[/tex]×x

Answer: Slope 1/2

y-int -5

graph (0,-5),(1,-9/2)

Step-by-step explanation:

Answer:

[tex]SE=\frac{1.5}{\sqrt{120}}=0.137[/tex]

The 95% confidence interval would be given by (4.729;5.271)    

Step-by-step explanation:

1) 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".

The standard error of a statistic is "the standard deviation of its sampling distribution or an estimate of that standard deviation"

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[/tex] represent the sample mean for the sample  

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

s represent the sample standard deviation

n represent the sample size  

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)

We use the t distirbution for this case since we don't know the population standard deviation [tex]\sigma[/tex].

Where the standard error is given by: [tex]SE=\frac{s}{\sqrt{n}}[/tex]

And the margin of error would be given by: [tex]ME=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

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=120-1=119[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. And we see that [tex]t_{\alpha/2}=1.98[/tex]

The standard error would be given by:

[tex]SE=\frac{1.5}{\sqrt{120}}=0.137[/tex]

Now we have everything in order to replace into formula (1) and calculate the interval:

[tex]5-1.98\frac{1.5}{\sqrt{120}}=4.729[/tex]    

[tex]5+1.98\frac{1.5}{\sqrt{120}}=5.271[/tex]

So on this case the 95% confidence interval would be given by (4.729;5.271)    

Final answer:

The standard error of the mean (SE) is calculated to be 0.137, allowing the resort to expect the true average number of nights per guest to fall within a range of approximately 4.73 to 5.27 nights with 95% confidence.

Explanation:

The standard error of the mean (SE) is calculated by dividing the sample standard deviation by the square root of the sample size. In this case, the standard error is 1.5 / sqrt(120) = 0.137. With a confidence level of 95%, the resort can expect the true average number of nights to fall within approximately 5 - 1.96 * 0.137 nights to 5 + 1.96 * 0.137 nights, which is roughly between 4.73 and 5.27 nights.

Answer:

0.212 is the probability that a randomly selected tire will have a depth less than 0.70 mm.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  0.98 mm

Standard Deviation, σ =  0.35 mm

We are given that the distribution of tire tread is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) P(depth less than 0.70 mm)

P(x < 0.70)

[tex]P( x < 0.70) = P( z < \displaystyle\frac{0.70 - 0.98}{0.35}) = P(z < -0.8)[/tex]

Calculating from normal z table, we have:

[tex]P(z<-0.8) = 0.212[/tex]

[tex]P(x < 0.70) = 0.212 = 21.2\%[/tex]

Thus, this event is not unusual and will  not warrant a refund.

Answer:

The number of ways is equal to [tex]12!8!20![/tex]

Step-by-step explanation:

The multiplication principle states that If a first experiment can happen in n1 ways, then a second experiment can happen in n2 ways ... and finally a i-experiment can happen in ni ways therefore the total ways in which the whole experiment can occur are

n1 x n2 x ... x ni

Also, given n-elements in which we want to put them in a row, the total ways to do this are n! that is n-factorial.

For example : We want to put 4 different objects in a row.

The total ways to do this are [tex]4!=4.3.2.1=24[/tex] ways.

Using the multiplication principle and the n-factorial number :

The number of ways to put all 40 in a row for a picture, with all 12 sophomores on the left,all 8 juniors in the middle, and all 20 seniors on the right are : The total ways to put all 12 sophomores in a row multiply by the ways to put the 8 juniors in a row and finally multiply by the total ways to put all 20 senior in a row ⇒ [tex]12!8!20![/tex]

There are 1.1657416 × 10^30 ways to arrange the students in a row with all the sophomores on the left, juniors in the middle, and seniors on the right.

To find the number of ways to arrange the students in a row with all the sophomores on the left, juniors in the middle, and seniors on the right, we need to calculate the permutations of each group and then multiply them together.

The number of ways to arrange the 12 sophomores is 12!, which is 479,001,600.

The number of ways to arrange the 8 juniors is 8!, which is 40,320.

The number of ways to arrange the 20 seniors is 20!, which is 2,432,902,008,176,640,000.

Multiplying these three numbers together, we get a total of 1.1657416 × 10^30 ways to arrange all the students in a row for a picture.

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Answer:

AAS method can be used to prove that the two triangles are congruent.

Step-by-step explanation:

According to the question for the two triangles one pair of opposite angles are equal. One another pair of angles are equal for the two and one pair of sides are also equal of the two.

Hence, the two given triangles are congruent by AAS rule.

Hence, AAS method can be used to prove that the two triangles are congruent.

Answer:

  see below

Step-by-step explanation:

Opposite sides are parallel in a parallelogram, so if they have different slope, the figure will not be a parallelogram.

_____

Comments on other answer choices

If the slope of diagonal EG is perpendicular to that of diagonal FH, it only proves the figure is some sort of kite. The figure may or may not be a parallelogram.

Adjacent sides being different lengths does not prove anything (except that the figure is not a rhombus).

Proving angle F is a right angle does not prove anything else about the shape of the figure. The figure may or may not be a parallelogram. (If it is a parallelogram, it is also a rectangle.)

Answer:

Have no fear, the answer is top left GIVE BRANLIEST PLZ

Step-by-step explanation:

Answer:

False this value is very likely to find in this distribution

Step-by-step explanation:

With mean 7 ( μ ) and standad deviation  (σ ) 2 we can observe, value 6.6  is close to the lower limit of the interval

μ  ± 0,5 σ      7 ± 1 in which we should find 68,3 % of all values

(just 6 tenth to the left)

And of course 6.6 is inside the interval

μ  ± 1 σ   where we find 95.7 % of th value

We conclude this value is not unlikely at all

Final answer:

To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we can use hypothesis testing. The calculated z-score is less than the critical z-value, indicating that it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in the population. Therefore, the statement is true.

Explanation:

To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we need to use hypothesis testing. We can set up the null and alternative hypotheses as follows:

Null hypothesis (H0): The population mean breaking weight is equal to 7 lbs.

Alternative hypothesis (Ha): The population mean breaking weight is less than 7 lbs.

Next, we can calculate the z-score using the formula (sample mean - population mean)/(standard deviation/sqrt(sample size)). Substituting the values, we get (6.6 - 7)/(2/sqrt(36)) = -0.6/ (2/6) = -0.6/ (1/3) = -1.8.

We can now look up the critical z-value for a one-tailed test at a significance level of 0.05. The critical z-value is -1.645. Since the calculated z-score (-1.8) is less than the critical z-value (-1.645), we can reject the null hypothesis. Therefore, it is true that finding a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs is very unlikely.

Answer:

82 apartments should be rented.

Maximum profit realized will be $30964.

Step-by-step explanation:

Monthly profit realized from renting out x apartments is modeled by

P(x) = -11x² + 1804x - 43000

To maximize the profit we will take the derivative of the function P(x) with respect to x and equate it to zero.

P'(x) = [tex]\frac{d}{dx}(-11x^{2}+1804x-43000)[/tex]

       = -22x + 1804

For P'(x) = 0,

-22x + 1804 = 0

22x = 1804

x = 82

Now we will take second derivative,

P"(x) = -22

(-) negative value of second derivative confirms that profit will be maximum if 82 apartments are rented.

For maximum profit,

P(82) = -11(82)² + 1804(82) - 43000

        = -73964 + 147928 - 43000

        = $30964

Therefore, maximum monthly profit will be $30964.

The number of units that should be rented out to maximize the monthly rental profit is 82 units.

The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]

To find the number of units that should be rented out to maximize the monthly rental profit and the maximum monthly profit, we need to analyze the given profit function:

[tex]\[ P(x) = -11x^2 + 1804x - 43,000 \][/tex]

This is a quadratic function of the form [tex]\( P(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = -11 \), \( b = 1804 \)[/tex], and c = -43,000 .

For a quadratic function [tex]\( ax^2 + bx + c \)[/tex] with a < 0  (which opens downwards), the maximum value occurs at the vertex. The x-coordinate of the vertex for the function P(x)  is given by:

[tex]\[ x = -\frac{b}{2a} \][/tex]

Substituting a = -11  and b = 1804 :

[tex]\[ x = -\frac{1804}{2(-11)} = \frac{1804}{22} = 82 \][/tex]

Therefore, the number of units that should be rented out to maximize the profit is x = 82 .

Next, we calculate the maximum monthly profit by substituting x = 82  back into the profit function:

[tex]\[ P(82) = -11(82)^2 + 1804(82) - 43,000 \][/tex]

Calculating step by step:

[tex]\[ 82^2 = 6724 \]\[ -11(6724) = -73,964 \]\[ 1804(82) = 147,928 \]\[ P(82) = -73,964 + 147,928 - 43,000 \]\[ P(82) = 30,964 \][/tex]

So, the maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]

In summary:

- The number of units that should be rented out to maximize the monthly rental profit is 82 units.

- The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]

Answer:

[tex]P(\bar X>308)=1-0.0721=0.928[/tex]

Step-by-step explanation:

1) 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".  

Let X the random variable that represent the GRE score at the University of Pennsylvania for the incoming class of 2016-2017, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu=311,\sigma=13)[/tex]  

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

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

On this case  [tex]\bar X \sim N(311,\frac{13}{\sqrt{40}})[/tex]

2) Calculate the probability

We want this probability:

[tex]P(\bar X>308)=1-P(\bar X<308)[/tex]

The best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

If we apply this formula to our probability we got this:

[tex]P(\bar X >308)=1-P(Z<\frac{308-311}{\frac{13}{\sqrt{40}}})=1-P(Z<-1.46)[/tex]

[tex]P(\bar X>308)=1-0.0721=0.9279[/tex]  and rounded would be 0.928

One sample t-test

Step-by-step explanation:

In this statistical test, you will be able to test if a sample mean, significantly differs from a hypothesized value.Here you can test if the average swimming velocity differs significantly from an identified value in the hypothesis.Then you can conclude whether the group of 18 fish or that of 21 fish has a significantly higher or lower  mean velocity than the one in the hypothesis.

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Answer:

5

Step-by-step explanation:

27-22= 22-17 = 17-12 =12-7=d

d= 5

The common difference of the arithmetic sequence 7, 12, 17, 22, 27,... is 5, which is the constant difference between consecutive terms of the sequence.

The sequence given is an arithmetic sequence, which is a sequence of numbers where the difference between consecutive terms is constant. To find the common difference in the arithmetic sequence 7, 12, 17, 22, 27,..., we subtract any term from the preceding term in the sequence. Let's perform this calculation using the first two terms:

Thus, the common difference is 5. We can confirm this by checking the difference between other consecutive terms:

All differences are the same, confirming that the common difference of the sequence is indeed 5.

On average, when placing a $1 bet on five numbers in roulette, you would expect to lose about 53 cents due to the game's probability structure.

In order to calculate the expected winnings in a roulette game, consider a $1 bet on five numbers. The roulette wheel has 38 possibilities (1 to 36, 0, and 00). So, the chance that your bet will win is 5 out of 38. If it wins, the game pays off 6 to 1, which means you get your $1 bet back and win $6 additional, for a total of $7. The expected value of this bet can be calculated as follows: (probability of winning * amount won if bet is successful) - (probability of losing * amount lost if bet is not successful). In other words, E(X) = [5/38 * $7] - [33/38 * $1]. This equates to an expected value of -$0.53 (rounded to the nearest cent). So, on average, with each $1 bet on five numbers, you would expect to lose about 53 cents.

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In the game of roulette with 38 outcomes, if you place a $1 bet on five numbers, your expected winnings are -$0.08, which means you lose 8 cents per game on average.

To calculate the expected winnings of a bet in this roulette scenario, you need to know the probability of winning and the associated payout, and also the probability of losing. In a roulette wheel with numbers 1-36 plus 0 and 00, there are a total of 38 possible outcomes. When you place a 5-number bet, the probability of winning is 5 out of 38 or approximately 0.1316, and the probability of losing is 33 out of 38 or approximately 0.8684.

Now we add the expected winnings and losses. The expected gain from a win is the probability of winning times the payout, which is $6 in this case. So, 0.1316 * $6 = $0.79, rounded to the nearest cent. The expected loss from a bet is the probability of loss times the amount of the bet which is $1. So, 0.8684 * -$1 = -$0.87, rounded to the nearest cent. Total expected value or earnings from a single $1 bet on five numbers would be the sum of expected gains and losses, or $0.79 - $0.87 = -$0.08.

So, with a $1 bet on five numbers on this roulette wheel, you can expect, on average, to lose 8 cents per game. This negative figure indicates that this is not a good bet if you're expecting to make money on average.

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Answer:

Step-by-step explanation:

To estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence, we need to survey at least 92 Clemson students.

To estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence, we need to calculate the sample size needed for this level of precision. The formula to determine the sample size is:

n = (z * σ / E)^2

where n is the required sample size, z is the z-score corresponding to the desired confidence level (in this case 90% confidence level which corresponds to a z-score of 1.645), σ is the standard deviation, and E is the desired margin of error (1.5 years).

Plugging in the values, we get:

n = (1.645 * 6 / 1.5)^2

n = 91.154

Rounding up to the nearest whole number, we need to survey at least 92 Clemson students in order to estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence.

Answer:

Lowest acceptable score = 121.3

Step-by-step explanation:

Mean test score (μ) = 115

Standard deviation (σ) = 12

The z-score for any given test score 'X' is defined as:  

[tex]z=\frac{X-\mu}{\sigma}[/tex]  

In this situation, the organization is looking for people who scored in the upper 30% range, that is, people at or above the 70-th percentile of the normally distributed scores. At the 70-th percentile, the corresponding z-score is 0.525 (obtained from a z-score table). The minimum score, X, that would enable a candidate to apply for membership is:

[tex]0.525=\frac{X-115}{12}\\X=121.3[/tex]

Answer:

[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158[/tex]

Step-by-step explanation:

The probability distribution of sampling distribution [tex]\hat{p}[/tex] is known as it sampling distribution.

The mean and standard deviation of the proportion is given by :-

[tex]\mu_{\hat{p}}=p\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex]

, where p =population proportion and  n= sample size.

Given : According to a survey, 50% of Americans were in 2005 satisfied with their job.

i.e. p = 50%=0.50

Now, for sample size n= 1000 , the mean and standard deviation of the proportion will be :-

[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{0.50(1-0.50)}{1000}}=\sqrt{0.00025}\\\\=0.0158113883008\approx0.0158[/tex]

Hence, the mean and standard deviation of the proportion for a sample of 1000:

[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158[/tex]

Answer: 95% confidence interval would be (175.04,194.96).

Step-by-step explanation:

Since we have given that

Mean = 185 mg

Standard deviation = 17.6 mg

At 95% confidence level, z = 1.96

So, Interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=185\pm 1.96\times \dfrac{17.6}{\sqrt{12}}\\\\=185\pm 9.958\\\\=(185-9.958,185+9.958)\\\\=(175.042,194.958)\\\\=(175.04,194.96)[/tex]

Hence, 95% confidence interval would be (175.04,194.96).

Using the provided values and steps for constructing a 95% confidence interval, we estimate, with 95% confidence, that the true mean cholesterol content of all such eggs lies between 146.5 and 223.5 milligrams.

To construct a 95% confidence interval for the mean cholesterol content of all such eggs, we will use the provided sample mean (185 milligrams), standard error (17.6 milligrams), and the fact that the sample size (12 eggs) is relatively small, so we use a t-distribution.

Firstly, we need to find the t-score for a 95% confidence level. Since degrees of freedom (df) is n-1 -> 12-1=11, and at a 95% confidence level, the t-score (from t-distribution table) is approximately 2.201 for a two-tailed test.

Next, we use the formula for confidence interval:

Calculating these gives:

So, we estimate with 95 percent confidence that the true mean cholesterol content of all such eggs is between 146.5 and 223.5 milligrams.

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Answer:

The length of the person’s shadow is 5.7ft

Explanation:

Length of the flagpole =a= 35ft

Length of the shadow of the flagpole= b=50ft

Length of the person=c= 4ft

Suppose the length of the person’s shadow is=d

According to the  rules of trigonometry

[tex]\frac{\text { Length of the flagpole }}{\text { Length of the shadow of the flagpole }}=\frac{\text { Length of the person }}{\text { Length of the person's shadow }}[/tex]

[tex]\frac{a}{b}=\frac{c}{d}[/tex]

[tex]\frac{35}{50}=\frac{4}{d}[/tex]

35d=200

d=[tex]\frac{200}{35}[/tex]

d=5.7ft  

Hence, The length of the person’s shadow is 5.7ft.