A sample of 161children was selected from fourth and fifth graders at elementary schools in Philadelphia. In addition to recording the grade level, the researchers determined whether each child had a previously undetected reading disability. Sixty-six children were diagnosed with a reading disability. Of these children, 32 were fourth graders and 34 were fifth graders. Similarly, of the 95 children with normal reading achievement, 55 were fourth graders and 40 were fifth graders.
a. Identify the two qualitative variables (and corresponding levels) measured in the study.
b. From the information provided, form a contigency table.
c. Assuming that the two variables are independent, calculate the expected cell counts.

Answer:

21.75 square units

Step-by-step explanation:

Draw a radial line from O to the point where AB intersects the circle.  We'll call this point P.

Draw another radial line from O to the point where arc BC intersects the circle.  We'll call this point Q.

OQ is equal to the radius of the circle, 2.  And AQ is equal to the radius of the sector, 8.  Therefore, the length of AO is 6.

Next, OP is equal to the radius of the circle, 2.  Since AB is tangent to the circle, it is perpendicular to OP.

So we have a right triangle with a hypotenuse of 6 and a short leg of 2.  Finding the angle ∠PAO:

sin ∠PAO = 2/6

∠PAO = asin(1/3)

That means the angle ∠BAC is double that:

∠BAC = 2 asin(1/3)

∠BAC ≈ 38.94°

Therefore, the area of the sector is:

A = (θ/360) πr²

A = (38.94/360) π(8)²

A ≈ 21.75

The area of the sector is approximately 21.75 square units.

Answer:

a) [tex]x=a\cdot cos(b), y=-a\cdot sin(b), 0 \leq b \leq 2\pi[/tex]

b) [tex]x=a\cdot cos(b), y=a\cdot sin(b), 0 \leq b \leq 2\pi[/tex]

c) [tex]x=a\cdot cos(b), y=-a\cdot sin(b), 0 \leq b \leq 6\pi[/tex]

d) [tex]x=a\cdot cos(b), y=a\cdot sin(b), 0 \leq b \leq 6\pi[/tex]

Step-by-step explanation:

The parametric equation for a circle is:

[tex]x=a\cdot cos(b)[/tex]

[tex]y=a\cdot sin(b)[/tex]

Where a is the radius and b is the angular displacement.

a) If a is negative in y and 0 ≤ b ≤ 2π, we have clockwise moves.

[tex]x=a\cdot cos(b), y=-a\cdot sin(b), 0 \leq b \leq 2\pi[/tex]

b)  If a is positive in y and 0 ≤ b ≤ 2π, we have counterclockwise moves.

[tex]x=a\cdot cos(b), y=a\cdot sin(b), 0 \leq b \leq 2\pi[/tex]

c) If a is negative in y and 0 ≤ b ≤ 6π, we have three times clockwise moves.

[tex]x=a\cdot cos(b), y=-a\cdot sin(b), 0 \leq b \leq 6\pi[/tex]

d) If a is positive in y and 0 ≤ b ≤ 6π, we have three times counterclockwise moves.

[tex]x=a\cdot cos(b), y=a\cdot sin(b), 0 \leq b \leq 6\pi[/tex]

Have a nice day!

The question requires finding parametric equations for a particle's motion along a circle of radius a, both clockwise and counterclockwise for one and three complete cycles. The equations are based on the standard circle parametric equations with adjustments to the angle parameter \(\theta\) and the interval according to the direction and number of cycles.

The student is asking for the parametric equations and parameter intervals for the motion of a particle tracing a circle, both clockwise and counterclockwise, once and three times. A circle with radius a and center at the origin has standard parametric equations x = a cos(\theta) and y = a sin(\theta), where \(\theta\) is the parameter usually representing an angle in radians.

Therefore, the appropriate parametric equations for each situation are:

Answer:

a. 20.23

b. 14.1

c. 8.3, 9.2, 9.8, 12.1, 13.5, 14.7, 26.5, 34.7, 36.5, 37

d. 22.65

Step-by-step explanation:

Sum = 202.3

Count = 10

Mean = 202.3 / 10

Mean = 20.23

Since there is an even number of data values the median is the mean of the two data values in the middle, calculated as follows;

8.3, 9.2, 9.8, 12.1, 13.5, 14.7, 26.5, 34.7, 36.5, 37

Median = (13.5 + 14.7) / 2

Median = 14.1

Mode is the value or values in the data set that occur most frequently. Since all are occurred once so all the values will be considered as Mode;

Mode = 8.3, 9.2, 9.8, 12.1, 13.5, 14.7, 26.5, 34.7, 36.5, 37

Midrange = (Highest Value + Lowest Value) / 2

Midrange = (37 + 8.3) / 2

Midrange = 22.65

Answer:

Paired  t test

Step-by-step explanation:

Given that you collect a random sample of 28 adult golfers and record two scores for each: one taken before the subject receives professional coaching and one taken after.

Here subject of interest is to study whether the professional coaching really improves the scores.

For this two groups are taken from golfers and they were given chances to play and scores were recorded before and after coaching.  The same subject with two different scores recorded and the difference calculated.  Hence here the appropriate test is paired t test

Statistic should be t because population std devition is not known.

Hence paired t test for comparison of mean differences before and after should be done.

Answer:

z-test statistic is 3.38

Step-by-step explanation:

Null and alternative hypotheses are:

[tex]H_{0}[/tex]: the proportion of papers that lacked statistical assistance  sent back without review is the same as the proportion of papers with statistical help sent back without review

[tex]H_{a}[/tex]: the proportion of papers that lacked statistical assistance sent back without review is different than the proportion of papers with statistical help sent back without review

Test statistic can be found using the equation:

[tex]z=\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where

Then [tex]z=\frac{0.71-0.57}{\sqrt{{0.61*0.39*(\frac{1}{190} +\frac{1}{514}) }}}[/tex] ≈ 3.38

Answer:

[tex]p = \frac{1564}{5020} = 0.3116[/tex]

Step-by-step explanation:

The point estimate for the population proportion is the number of sucesses divided by the size of the sample.

In this problem, we have that:

A success is being a tradition hogan in the reservation(the population). In a sample of 5020, 1564 are traditional hogans.

So [tex]p = \frac{1564}{5020} = 0.3116[/tex]

Answer:

0.001%

Step-by-step explanation:

Given that a recent study conducted by the state government attempts to determine whether the voting public supports a further increase in cigarette taxes to help the public schools

n = 1500

favour x = 622

Sample proportion p = [tex]622/1500 = 0.4147[/tex]

[tex]H_0: p = 0.66\\H_a: p >0.66[/tex]

(right tailed test)

Assume H0 to be true.

Std error = [tex]\sqrt{\frac{0.66*0.34}{1500} } \\=0.01223[/tex]

p difference = -0.2453

Test statistic Z = [tex]\frac{-0.2453}{0.01223} \\=-20.05[/tex]

p value <0.00001

Hence fon ull hypothesis not to be rejected significant level should be greater than 0.00001 or 0.001%

To construct a confidence interval for the difference between two proportions, you use an unbiased estimate of the difference between the two proportions.

When constructing a confidence interval for the difference between two proportions, an unbiased estimate of the difference between the two proportions is obtained using the formula:

(p1 - p2) ± ME

Where:

The margin of error is calculated as:

ME = z * sqrt((p1*(1-p1)/n1) + (p2*(1-p2)/n2))

Where:

Using this formula, you can calculate the confidence interval for the difference between two proportions.

Answer: The point estimate of the population mean is 25.

The margin of error for the confidence interval is 3 .

Step-by-step explanation:

The confidence interval for population mean is given by :-

[tex](\overline{x}-E , \overline{x}+E)[/tex] , here [tex]\overline{x}[/tex] is the point estimate of the population mean and E is the margin of error .

As per given , we have

Lower bound of CI = [tex]\overline{x}-E =22[/tex]   (1)

Upper bound of CI =  [tex]\overline{x}+E =28[/tex]  (2)

Add (1) and (2) , we get

[tex]2\overline{x}=50\\\Rightarrow\ \overline{x}=25[/tex]

Subtract (1) from (2) , we get

[tex]2E=6\\\Rightarrow\ E=3[/tex]

Hence, the point estimate of the population mean is 25.

The margin of error for the confidence interval is 3 .

Answer:

1. The point estimate for population mean is 25.

2)

[tex]\text{Margin of error} = \pm 3[/tex]

Step-by-step explanation:

We are given the following information in the question:

Confidence interval: (22,28)

Confidence interval is calculated as:

[tex]\text{Sample mean }\pm \text{ Margin of error}[/tex]

Thus, we can write the equations:

[tex]\bar{x} - \text{Margin of error} = 22\\\bar{x} + \text{Margin of error} = 28[/tex]

1) The point estimate of the population mean

To calculate the point estimate of the population mean we solve the two equations, to find the sample mean

Adding the two equations we get:

[tex]2\bar{x} = 22+ 28 = 50\\\bar{x} = 25[/tex]

Thus, the point estimate for population mean is 25.

2) The margin of error for the confidence interval

Putting the values from the equation, we get:

[tex]\text{Margin of error} = 28 - 25 = 3[/tex]

Thus, the margin of error f the given confidence interval is

[tex]\text{Margin of error} = \pm 3[/tex]

Answer:

a) Q = 122 units/order

b) Number of orders = 2.05 orders/year

c) Average inventory = 61 units

d) Ordering costs = 125 $/order

Step-by-step explanation:

The economic quantity order (EOQ) formula allow us to minimize the ordering cost, in function of the demand, ordering cost and holding cost.

The EOQ formula is:

[tex]EOQ=\sqrt{\frac{2DS}{H} }[/tex]

where:

D: demand in units/year

S: Order costs, per order

H: holding or carrying cost, per unit a year

a) In this case:

D: 250 u/year

S: 30 $/order

H: 1 $/year-unit

[tex]EOQ=\sqrt{\frac{2DS}{H} }=\sqrt{\frac{2*250*30}{1} }=\sqrt{15000}=122.47\approx122[/tex]

b) If we have a demand of 250 units/year and we place orders of 122 units, the amount of orders/year is:

[tex]\#orders=\frac{D}{EOQ}=\frac{250\,units/year}{122\,units/order}=2.05\, \frac{orders}{year}[/tex]

c) We assume that there is no safety stock, so everytime the stock hits 0 units, a new order enter the inventory.

In this case, the average inventory can be estimated as the average between the inventory when a new order enters the inventory (122 u.) and the inventory right before a order enters (0 u.)

[tex]\#av.inventory=\frac{122+0}{2}=61[/tex]

The average inventory is 61 units.

d) If 250 units is the optimal quantity for an order, it means it is equal to the EOQ. We can calculate the new ordering costs as:

[tex]EOQ=\sqrt{\frac{2DS}{H} }=\sqrt{\frac{2*250*S}{1} }=250\\\\2*250*S=250^2\\\\S=250/2=125\,\$/order[/tex]

A) To minimize cost, the number of units that should be ordered each time an order is placed is; 122 units

B) The number of orders per year needed with the optimal policy is; 2 orders per year.

C)  The average inventory if costs are minimized is; 61 units

D) For the order policy of Q = 250 to be optimal, the ordering cost would have to be; $125 per order

The formula for economic quantity order (EOQ) is given as;

EOQ = √(2DS/H)

Where;

We are given;

Annual demand; D = 250 units/year

Holding Cost; H = $1 per unit per year

Set up costs; S = $30 per order

EOQ = √(2 × 250 × 30/1)

EOQ = 122.47

But EOQ has to be a whole number and so we approximate to the nearest whole number to get;

EOQ = 122

n = D/EOQ

Plugging in the relevant values gives;

n = 250/122

n = 2.049

But number of orders has to be a whole number. Thus, we approximate to the nearest whole number to get;

n = 2 orders per year

Before a new order enters the inventory is 0 if we assume that there is no safe stock. Thus;

average inventory = (122 + 0)/2 = 61 units

EOQ = √(2DS/H)

⇒ 250 = √(2 × 250 × S/1)

Square both sides to get;

250² = 500S

S = 250²/500

S = $125 per order

Read more about economic quantity order (EOQ) at; https://brainly.com/question/16395657

Answer:

B. Rational functions

Step-by-step explanation:

The partial fraction decomposition is used for functions there are described by fractions, and for which the substitution method is not possible. These are rational functions, in which both the numerator and the denominator are polynomials.

So the correct answer is:

B. Rational functions

Partial fraction decomposition is used to simplify complex rational functions to make them easier to integrate. Rational function is the correct answer.

The correct answer is B. Rational functions. Partial fraction decomposition is a mathematical technique used essentially to simplify complex rational functions. A rational function is a function that can be defined as the ratio of two polynomials. By breaking down a complex rational function into simpler fractions (which is what partial fraction decomposition does), integration becomes more manageable. For example, it's easier to integrate simple fractions like 1/x or 2/x^2, which would be the result of a partial fraction decomposition, than complex, intertwined expressions.

https://brainly.com/question/34850694

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Answer:

a) [tex]y = 3.8 x +100[/tex]

b) [tex]Abs. change= |168.4-167.5|=0.9[/tex]

So the calculated value is 0.9 points above the actual value.

[tex]Relative. Change =\frac{|168.4 -167.5|}{167.5}x100 =0.537[/tex]%

And the calculated value it's 0.537% higher than the actual value.

c) For this case we can use the slope obtained from the linear model to answer this question, and we can conclude that the CPI is increasing at approximate 3.8 units per year.

Step-by-step explanation:

Data given

1982 , CPI=100

1986, CPI = 191.2

Notation

Let CPI the dependent variable y. And the time th independent variable x.

For this case we want to adjust a linear model givn by the following expression:

[tex]y=mx+b[/tex]

Solution to the problem

Part a

For this case we can find the slope with the following formula:

[tex] m =\frac{CPI_{2006}-CPI_{1982}}{2006-1982}[/tex]

And if we replace we got:

[tex] m =\frac{191.2-100}{2006-1982}=3.8[/tex]

Let X represent the number of years after. Then for 1982 t = 0, and if we replace we can find b:

[tex] 100 = 3.8(0)+b[/tex]

And then [tex]b=100[/tex]

So then our linear model is given by:

[tex]y = 3.8 x +100[/tex]

Part b

For this case we need to find the years since 1982 and we got x = 2000-1982=18, and if we rpelace this into our linear model we got:

[tex]y = 3.8(18) +100=168.4[/tex]

And the actual value is 167.5 we can compare the result using absolute change or relative change like this:

[tex]Abs. change= |168.4-167.5|=0.9[/tex]

So the calculated value is 0.9 points above the actual value.

And we can find also the relative change like this:

[tex]Relative. Change =\frac{|Calculated -Real|}{Real}x100[/tex]

And if we replace we got:

[tex]Relative. Change =\frac{|168.4 -167.5|}{167.5}x100 =0.537[/tex]%

And the calculated value it's 0.537% higher than the actual value.

Part c

For this case we can use the slope obtained from the linear model to answer this question, and we can conclude that the CPI is increasing at approximate 3.8 units per year.

Answer:

Experimental smoothing forecast for the next period will be 58.9

Step-by-step explanation:

We have given actual demand is [tex]A_{t-1}=61[/tex]

Initial forecast [tex]f_{t-1}=61[/tex] and [tex]\alpha =0.3[/tex]

We have to find the experimental smoothing forecast [tex]f_t[/tex]

Experimental forecast is given by

[tex]f_t=(1-\alpha )f_{t-1}+\alpha A_{t-1}=(1-0.3)\times 58+0.3\times 61=58.9[/tex]

So experimental smoothing forecast for the next period will be 58.9

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

[tex]n_1 [/tex] represent the sample after the preparation

[tex]n_2 [/tex] represent the sample without preparation  

[tex]\bar x_1 =678[/tex] represent the mean sample after preparation

[tex]\bar x_2 =1837[/tex] represent the mean sample without preparation

[tex]s_1 =197[/tex] represent the sample deviation after preparation

[tex]s_2 =328[/tex] represent the sample deviation without preparation

[tex]\alpha=0.1[/tex] represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex]

The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2[/tex]

Since the Confidence is 0.90 or 90%, 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. The excel command would be: "=-T.INV(0.025,df)  

The standard error is given by the following formula:  

[tex]SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}[/tex]  

After replace in the formula for the confidence interval we got this:

[tex]3.0 < \mu_1 -\mu_2 <13.0 [/tex]

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Answer:

Equation: [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]

Intersection: [tex](y-2)^2+(z-5)^2=7[/tex]

Step-by-step explanation:        

We are asked to write an equation of the sphere with center center [tex](-3,2,5)[/tex] and radius 4.    

We know that equation of a sphere with radius 'r' and center at [tex](h,k,l)[/tex] is in form:

[tex](x-h)^2+(y-k)^2+(z-l)^2=r^2[/tex]

Since center of the given sphere is at point [tex](-3,2,5)[/tex], so we will substitute [tex]h=-3[/tex], [tex]k=2[/tex], [tex]l=5[/tex] and [tex]r=4[/tex] in above equation as:

[tex](x-(-3))^2+(y-2)^2+(z-5)^2=4^2[/tex]

[tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]

Therefore, our required equation would be [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex].

To find the intersection of our sphere with the yz-plane, we will substitute [tex]x=0[/tex] in our equation as:

[tex](0+3)^2+(y-2)^2+(z-5)^2=16[/tex]

[tex]9+(y-2)^2+(z-5)^2=16[/tex]

[tex]9-9+(y-2)^2+(z-5)^2=16-9[/tex]

[tex](y-2)^2+(z-5)^2=7[/tex]

Therefore, the intersection of the given sphere with the yz-plane would be [tex](y-2)^2+(z-5)^2=7[/tex].

Answer:both students are incorrect

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)

Slope = (y2 - y1)/(x2 - x1)

y2 = 1

y1 = 4

x2 = 6

x1 = - 3

Slope = (1 - 4)/(6 - -3) = -3/9 = -1/3

To determine the intercept, we would substitute m = -1/3, x = 6 and y = 1 into y = mx + c. It becomes

1 = -1/3 × 6 + c = -2 + c

c = 1 + 2 = 3

The equation becomes

y = -x/3 + 3

If the equation was written in the slope intercept form which is expressed as

y - y1 = m(x - x1)

It becomes

y - 4 = -1/3(x - - 3)

y - 4 = -1/3(x + 3)

Both students are incorrect

Answer:

3.573 to 4.127

Step-by-step explanation:

Given

Sample size = 15

Mean = Sum of ratings/ sample size

Mean = 57.7/15

Mean = 3.85

Degree of freedom = sample size - 1

Degree of freedom = 15 - 1 = 14

df = 14

Then we calculate the standard deviation

(x - mean)² ||

(3.6 - 3.85)² || 0.0625

(2.9 - 3.85)² || 0.9025

(3.8 - 3.85)² || 0.0025

(4.5 - 3.85)² || 0.4225

(3.2 - 3.85)² || 0.4225

( 3.9 - 3.85)² || 0.0025

( 3.3 - 3.85)² || 0.3025

( 4.6 - 3.85)² || 0.5625

(4.1 - 3.85)² || 0.0625

(4.3 - 3.85)² || 0.2025

4.4 - 3.85)² || 0.3025

( 3.9 - 3.85)² || 0.0025

(3.2 - 3.85)² || 0.4225

( 4.2 - 3.85)² || 0.1225

( 3.8 - 3.85)² || 0.0025

Total || 3.7975

Variance = 3.7975/15 = 0.253167

Standard Deviation = √0.253167 = 0.50315703314174194

Standard Deviation = 0.5 ------- Approximated

The next step is to subtract the confidence level from 1, then divide by two.

i.e (1 - 0.95)/2 = 0.025

α = 0.025

Then we look up this answer to step in the t-distribution table.

For 14 degrees of freedom (df) and α = 0.025, my result is 2.145

The next step is to divide the sample standard deviation by the square root of the sample size.

0.5 / √15 = 0.129

Next is to multiply this result by step 2.145 (from the t table)

0.129 × 2.45 = 0.277

For the lower end of the range, subtract 0.277 from the sample mean.

3.85 – 0.277 = 3.573

Step 7: For the upper end of the range, add step 0.277 to the sample mean.

3.85 + 0.277 = 4.127

Answer:

a) Null hypothesis:[tex]\mu_{1} \leq \mu_{2}[/tex]

Alternative hypothesis:[tex]\mu_{1} > \mu_{2}[/tex]

b) [tex]z_{crit}=2.33[/tex]

And since the calculated value is lower than the critical value we have enough evidence at 0.01 of significance to FAIL to reject the null hypothesis.

Step-by-step explanation:

Part a

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

We want to test this:

Null hypothesis:[tex]\mu_{1} \leq \mu_{2}[/tex]

Alternative hypothesis:[tex]\mu_{1} > \mu_{2}[/tex]

Part b

Golf course designer Roberto Langabeer is evaluating two sites, Palmetto Dunes and Ocean Greens, for his next golf course. He wants to prove that Palmetto Dunes residents (population 1) play golf more often than Ocean Greens residents (population 2). Roberto commissions a market survey to test this hypothesis. The market researcher used a random sample of 64 individuals from each suburb, and reported the following:  X 1 = 15  times per month and  X 2 = 14  times per month. Assume that  σ 1 = 2  and  σ 2 = 3 . With  α = .01 , the critical z value is _____.

Data given and notation

[tex]\bar X_{1}=15[/tex] represent the mean for the sample 1

[tex]\bar X_{2}=14[/tex] represent the mean for the sample 2

[tex]\sigma_{1}=2[/tex] represent the population deviation for 1

[tex]\sigma_{2}=3[/tex] represent the population deviation for 2

[tex]n_{1}=64[/tex] sample size selected for 1

[tex]n_{2}=64[/tex] sample size selected for 2

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

The statistic is given by:

[tex]z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex]     (1)

[tex]z=\frac{15-14}{\sqrt{\frac{2^2}{64}+\frac{3^2}{64}}}}=2.22[/tex]  

In order to find the critical value we need a value that accumulates 0.01 of the area on the right tail, since we are conducting a right tailed test. And the critical value is:

[tex]z_{crit}=2.33[/tex]

And since the calculated value is lower than the critical value we have enough evidence at 0.01 of significance to FAIL to reject the null hypothesis.

Answer:

a) The 90% confidence interval would be given by (71.689;78.311)  

The 95% confidence interval would be given by (71.012;78.988)  

b) [tex]p_v =2*P(t_{24}<-2.583)=0.016[/tex]  

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.  So we have a significant effect.

Step-by-step explanation:

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".  

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".

[tex]\bar X=75[/tex] represent the sample mean  

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

[tex]s=9.68[/tex] represent the sample standard deviation  

n=25 represent the sample size  

Assuming the X follows a normal distribution  

[tex]X \sim N(\mu, \sigma=0.15[/tex]

The sample mean [tex]\bar X[/tex] is distributed on this way:

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

Part a

The confidence interval on this case is given by:

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

We can find the degrees of freedom like this:

[tex]df=n-1=25-1=24[/tex]

confidence 90%

The next step would be find the value of [tex]\t_{\alpha/2}[/tex], [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex]  

Using the t distribution with 24 df, excel or a calculator we see that:  

[tex]t_{\alpha/2}=1.71[/tex]

Since we have all the values we can replace:

[tex]75 - 1.71\frac{9.68}{\sqrt{25}}=71.689[/tex]  

[tex]75 + 1.71\frac{9.68}{\sqrt{25}}=78.311[/tex]  

So on this case the 90% confidence interval would be given by (71.689;78.311)  

Confidence 95%

The next step would be find the value of [tex]\t_{\alpha/2}[/tex], [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex]  

Using the t distribution with 24 df, excel or a calculator we see that:  

[tex]t_{\alpha/2}=2.06[/tex]

Since we have all the values we can replace:

[tex]75 - 2.06\frac{9.68}{\sqrt{25}}=71.012[/tex]  

[tex]75 + 2.06\frac{9.68}{\sqrt{25}}=78.988[/tex]  

So on this case the 95% confidence interval would be given by (71.012;78.988)  

Part b

Null hypothesis:[tex]\mu = 80[/tex]  

Alternative hypothesis:[tex]\mu \neq 80[/tex]  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]t=\frac{75-80}{\frac{9.68}{\sqrt{25}}}=-2.583[/tex]  

P-value  

The degrees of freedom are 25-1=24

Since is a two tailed test the p value would given by:  

[tex]p_v =2*P(t_{24}<-2.583)=0.016[/tex]  

Conclusion  

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.  So we have a significant effect.

Answer:

All of them are polynomial functions

Step-by-step explanation:

Remember that a polynomial function of x is a function whose value f(x) is always equal to [tex]f(x)=a_0+a_1x+a_2x^2+\cdots a_nx^n[/tex] for a fixed n≥0 (the degree of f) and fixed coefficients [tex]a_i\in\mathbb{R}[/tex]

For example, [tex]f(x)=x^2+3x[/tex] is a polynomial function, but [tex]g(x)=2^x+x[/tex] is not because [tex]2^n[/tex] is not a nonnegative power of x. Another example of a non-polynomial function is [tex]g(x)=x^{-1}=\frac{1}{x}[/tex].

f(x)=4⋅11x is polynomial with degree 1 and [tex]a_0=0,a_1=4\cdot 11[/tex]. For the same reasons, f(x)=3⋅18x and f(x)=10⋅17x are polynomial functions.

f(x)=−4x³−4x²+5x+1 is a polynomial function of degree 3 with [tex]a_0=1,a_1=5, a_2=a_3=-4[/tex]. and f(x)=−2x−1 is a polynomial function of degree 1 and coefficients [tex]a_0=-1,a_1=-2[/tex].

In mathematics, a polynomial function involves operations of addition, subtraction, multiplication, and non-negative, whole-number exponents of variables. Here, the polynomial functions from the given list are: f(x)=−4x3−4x2+5x+1 and f(x)=−2x−1.

In the realm of mathematics, a polynomial function is an expression consisting of variables and coefficients, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents of variables. When asked to determine which of the given functions are polynomial functions, we can apply this definition. Your options were: f(x)=4⋅11x, f(x)=3⋅18x, f(x)=10⋅17x, f(x)=−4x3−4x2+5x+1, and f(x)=−2x−1.

By our definition, the polynomial functions from the list are:  f(x)=−4x3−4x2+5x+1 and f(x)=−2x−1. The other three functions are not polynomial functions as they do not comply with the characteristics of polynomial functions.

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Answer:

Option a is right

Step-by-step explanation:

Given that a random sample of 250 men yielded 175 who said they'd ridden a motorcycle at some time in their lives, while a similar sample of 215 women yielded only 43 that had done so.

For proportions since binomial and sample size large we can use  z critical values.

Sample             I            II

N                    250        215     465

X                     175           43    218

p                       0.7          0.2    0.4688

p difference = 0.5

Std error of difference = [tex]\sqrt{p(1-p)(\frac{1}{n_1}+  \frac{1}{n_2} }\\=\sqrt{0.4688*0.5312)(\frac{1}{250} + \frac{1}{215} )}\\=0.0409[/tex]

Margin of error for 99% = 2.58*std error =  0.105

Confidence interval 99% = (0.5±0.105)            

Option a is right.

Substituting [tex]z[/tex] from the cone's equation,

[tex]z=\dfrac13\sqrt{x^2+y^2}[/tex]

into the equation of the sphere,

[tex]x^2+y^2+z^2=4[/tex]

gives the intersection of the two surfaces,

[tex]x^2+y^2+\left(\dfrac13\sqrt{x^2+y^2}\right)^2=4\implies x^2+y^2=\dfrac{18}5[/tex]

which is a circle of radius [tex]\sqrt{\frac{18}5}[/tex] centered at [tex]\left(0,0,\frac13\sqrt{\frac{18}5}\right)[/tex].

We parameterize this part of the sphere outside the cone (call it [tex]S[/tex]) by

[tex]\vec s(u,v)=\langle2\cos u\sin v,2\sin u\sin v,2\cos v\rangle[/tex]

with [tex]0\le u\le2\pi[/tex] and [tex]\cos^{-1}\frac1{\sqrt{10}}\le v\le\pi[/tex].

Take the normal vector to [tex]S[/tex] to be

[tex]\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=\langle4\cos u\sin^2v,4\sin u\sin^2v,4\cos v\sin v\rangle[/tex]

Then the area of [tex]S[/tex] is

[tex]\displaystyle\iint_S\mathrm dA=\int_0^{2\pi}\int_{\cos^{-1}\frac1{\sqrt{10}}}^\pi\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|\,\mathrm dv\,\mathrm du[/tex]

[tex]=\displaystyle2\pi\int_{\cos^{-1}\frac1{\sqrt{10}}}^\pi4\sin v\,\mathrm dv=\boxed{\frac{40+4\sqrt{10}}5\pi}[/tex]

Answer:

$6

Explanation:

The standard deviation gives an idea of the dispertion of values in those 12 month. A hight value of standard deviation means that the prices changed in big increments ( for example one month is $20, other month is $60 and other $85). If that  would be the case, is important to investors to know it and the deviation should be reported.

So, given that it isnt reported, we can say that the price vary but only a few dollars, resulting in a small standard deviation such as $6.

Final answer:

A standard deviation of $6 for a blue chip stock is more likely than $26 or $60, as blue chip stocks are typically stable with less price volatility.

Explanation:

Regarding the standard deviation of the blue chip stock price, it is unlikely that the standard deviation would be very high considering the nature of blue chip stocks. Blue chip stocks are known for their stability and are typically less volatile than other types of stocks. Therefore, a standard deviation of $60 would suggest a very high level of volatility, which is uncharacteristic of blue chip stocks. A standard deviation of $6 is more plausible because it indicates minor fluctuations around the average price of $72, which is more in line with the expected behavior of blue chip stocks. A standard deviation of $26 is possible but less likely than $6, given that such a value still represents a fairly high level of volatility for blue chip stocks, which are commonly considered as safe, long-term investments with relatively steady prices.

Answer:

93

Step-by-step explanation:

Using long division, this is the correct answer. Here are the steps:

remove decimal points: 744 ÷ 8

divide using long division: 93

Answer:

93

Step-by-step explanation:

Given number = 74.4

Divisor = 0.8

[tex]\[\frac{74.4}{0.8} = \frac{74.4*10}{0.8*10}\][/tex]

[tex]\[= \frac{744}{8}\][/tex]

Simplifying,

[tex]\[= \frac{744\div 4}{8\div 4}\][/tex]

[tex]\[= \frac{186}{2}\][/tex]

[tex]\[= 93\][/tex]

Validating by multiplying the quotient and divisor,

[tex]\[93 * 0.8\][/tex]

[tex]\[= 74.4\][/tex]

This is equal to the dividend.

Answer: 0.6

Step-by-step explanation:

Formula to get the standardized value :

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]  (1)

, where x= random variable , [tex]\mu[/tex] = Population mean and [tex]\sigma[/tex] = population standard deviation.

As per given , we have

[tex]\mu=25[/tex]

[tex]\sigma=5[/tex]

To find standardized value for the score of x=28

[tex]z=\dfrac{28-25}{5}=0.6[/tex]   (substitute values in (1))

Hence, the standardized value for the score of 28 is 0.6 .

Answer:

D. Discrete

Step-by-step explanation:

The number of reported car accidents is a countable variable, that can only be natural values.

You can have 0, 1, 2, 3, 4, ..., 100, ..., 1000 reported car accidents in a given month.

You cannnot have 4.5 reported car accidents in a given month, for example.

So the correct answer is:

D. Discrete

The type of variable that represents the number of auto accidents reported in a given month is a discrete variable, as it can only take certain values (0, 1, 2, etc.), and never a fraction or decimal.

The type of variable that best describes the number of auto accidents reported in a given month is Discrete.

In statistics, variables are classified into different types based on their properties. A discrete variable is one that can only take certain values. This is the case for the number of auto accidents in a month, which could be 0, 1, 2, 3, etc., but never a fraction or decimal value like 2.5.

In contrast, a continuous variable could take any value within a certain range, including fractions and decimals. Meanwhile, interval and ratio are types of data measurements that don't apply in this context.

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Sampling is most appropriate

option c

In the given multiple choice situations, the use of sampling would most probably be appropriate in the situation where 'The need for precise information is less important'. Sampling is a procedure of statistical analysis where a predetermined number of observations are taken from a larger population. It is typically used when it is impracticable or expensive to study the entire population. It's an ideal method when having precise and exact information is not as crucial, however, the results will be sufficient to make accurate predictions or conclusions about a larger group.

On the other hand, sampling may not be substantially beneficial for smaller populations or where the likelihood of selecting a representative sample is low because the results might be skewed or non-representative of the larger group. In such conditions, examining the entire population or utilizing another form of data collection would be more advantageous.

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Answer:

(D) 21

Explanation:

3(2x−1)−10=8+5x

Step 1: Simplify both sides of the equation.

3(2x−1)−10=8+5x

(3)(2x)+(3)(−1)+−10=8+5x(Distribute)

6x+−3+−10=8+5x

(6x)+(−3+−10)=5x+8(Combine Like Terms)

6x+−13=5x+8

6x−13=5x+8

Step 2: Subtract 5x from both sides.

6x−13−5x=5x+8−5x

x−13=8

Step 3: Add 13 to both sides.

x−13+13=8+13

x=21

Answer:

D. 21

Step-by-step explanation:

Given equation: [tex]\[3(2x – 1) – 10 = 8 + 5x\] [/tex]

Simplifying:

[tex]\[6x – 3 – 10 = 8 + 5x\] [/tex]

=> [tex]\[6x – 13 = 8 + 5x\] [/tex]

=> [tex]\[6x – 5x = 8 + 13\] [/tex]

=> [tex]\[(6 – 5)x = 21\] [/tex]

=> [tex]\[x = 21\] [/tex]

Verifying by substituting the value of x in the equation:

Left Hand Side: [tex]\[3(2x – 1) – 10\][/tex]

[tex]\[3(2*21 – 1) – 10\][/tex]

= [tex]\[3(42 – 1) – 10\][/tex]

= [tex]\[3(41) – 10\][/tex]

= [tex]\[123 – 10\][/tex]

= [tex]\[113\][/tex]

Right Hand Side: [tex]\[8 + 5x\][/tex]

= [tex]\[8 + 5*21\][/tex]

= [tex]\[8 + 105\][/tex]

= [tex]\[113\][/tex]

So, Left Hand Side = Right Hand Side when x=21.

Answer:

(7, 8)

Step-by-step explanation:

If Sidney places each piece 3 units east and 2 units north of the last one, we can figure out which coordinate comes first by subtracting the position of the second piece (S) by the position of the first piece (F):

[tex]S - F = (4,6) - (1,4) = (3,2)[/tex]

We can conclude that coordinates are in the (East, North) format.

Therefore, the location of the third piece (T) is:

[tex]T = S(E,N) +(3,2)= (4,6)+(3,2)\\T = (7,8)[/tex]

Answer:

Option D) The prisoner is actually guilty and the jury sets him free.

Step-by-step explanation:

We are given the following in the question:

Null Hypothesis:

The null hypothesis states that the prisoner is innocent

Alternate Hypothesis:

The alternate hypothesis states that the prisoner is guilty and not innocent.

Type II error:

It is the type error made when we fail to reject the ll hypothesis when it is actually false.

That is we accept a false null hypothesis.

Thus a type II error in the above scenario will be to accept that the prisoner is innocent (accepting the null hypothesis) when actually he is guilty( the alternate hypothesis)

Thus, type II error would be setting free a guilty prisoner.

Option D) The prisoner is actually guilty and the jury sets him free.

"The prisoner is actually guilty and the jury sets him free" represents the risk involved in making a Type II error. The correct option is d) The prisoner is actually guilty and the jury sets him free.

In statistical hypothesis testing, particularly in the context of a trial, we have two competing hypotheses:

Null Hypothesis (H₀): The prisoner is innocent (presumed innocent until proven guilty).

Alternative Hypothesis (H₁): The prisoner is guilty.

A Type II error, in this scenario, occurs when the jury fails to reject the null hypothesis (fails to convict the prisoner) when the alternative hypothesis is true (the prisoner is actually guilty).

Let's analyze the options in the context of Type II error:

(a) The prisoner is actually guilty and the jury sends him to jail. This is not a Type II error. This describes a correct decision where the jury convicts the guilty prisoner.

(b) The prisoner is actually innocent and the jury sends him to jail. This is a Type I error, where the jury convicts an innocent person (rejects the null hypothesis when it is true).

(c) The prisoner is actually innocent and the jury sets him free. This is the correct outcome when the jury correctly fails to reject the null hypothesis (the prisoner is innocent).

(d) The prisoner is actually guilty and the jury sets him free. This is a Type II error. Here, the jury fails to reject the null hypothesis (prisoner is innocent) when it is false (prisoner is guilty).

(e) The prisoner is sent to jail. This could happen in both correct decisions (prisoner is guilty) and Type I errors (prisoner is innocent but convicted).

Therefore, the answer that represents the risk involved in making a Type II error is d. The prisoner is actually guilty and the jury sets him free.