Write the quadratic equation whose roots are -2 and 1, and whose leading coefficient is 4 .

Answer:

The 95% confidence interval would be given by (14444.04;33657.30)

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".

Data: $26,650 $6,060 $52,820 $8,490 $13,660$25,840 $49,840 $23,790 $51,480 $18,960$990 $11,450 $41,810 $21,060 $7,860

We can calculate the mean and the deviation from these data with the following formulas:

[tex]\bar X= \frac{\sum_{i=1}^n x_i}{n}[/tex]

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

[tex]\bar X=24050.67[/tex] represent the sample mean for the sample  

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

s=17386.13 represent the sample standard deviation

n=15 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)

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=15-1=14[/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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,14)".And we see that [tex]t_{\alpha/2}=2.14[/tex]

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

[tex]24050.67-2.14\frac{17386.13}{\sqrt{15}}=14444.04[/tex]    

[tex]24050.67+2.14\frac{17386.13}{\sqrt{15}}=33657.30[/tex]

So on this case the 95% confidence interval would be given by (14444.04;33657.30)    

Answer: Margin of error = 6.58

Confidence interval =  (26.91, 40.08)

Step-by-step explanation:

Since we have given that

Sample size n = 8

Sample mean = 33.5 minutes

Population standard deviation = 9.5 minutes

At 95% confidence interval,

α = 0.05

t = 1.96

So, Margin of error is given by

[tex]t\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{9.5}{\sqrt{8}}\\\\=6.58[/tex]

Confidence interval would be

Lower limit:

[tex]\bar{x}-6.58\\\\=33.5-6.58\\\\=26.91[/tex]

Upper limit:

[tex]\bar{x}+6.58\\\\=33.5+6.58\\\\=40.08[/tex]

Hence, the interval would be (26.91, 40.08)

But at standard deviation 7.2 minutes, the confidence interval was (27.5, 39.5)

Confidence interval using the standard normal distribution is wider than the confidence interval using t distribution.

The information regarding the statistics shows that the margin of error is 6.58.

From the information given, the following can be deduced:

Therefore, the margin of error will be:

= 1.96 × 9.5/✓8

= 6.58

Also, the lower limit will be:

= 33.5 - 6.58

= 26.91

The upper limit will be:

= 33.5 + 6.58

= 40.08

Therefore, the interval will be (26.91, 40.08)

When the standard deviation is 7.2 minutes, the confidence interval is (27.5, 39.5).

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Answer:

z=0.930

[tex]p_v =P(z>0.930)=1-P(z<0.930)=1-0.824=0.176[/tex]

Step-by-step explanation:

1) Data given and notation    

[tex]\bar X=1165[/tex] represent the mean for the account balances of a credit company

[tex]s=125[/tex] represent the population standard deviation for the sample    

[tex]n=60[/tex] sample size    

[tex]\mu_o =1150[/tex] represent the value that we want to test  

[tex]\alpha[/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)

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean for account balances of a credit company is greater than 1150, the system of hypothesis would be:    

Null hypothesis:[tex]\mu \leq 1150[/tex]    

Alternative hypothesis:[tex]\mu > 1150[/tex]    

We don't know the population deviation, but the problem says the the distribution for the random variable is normal, so for this case we can use the z test to compare the actual mean to the reference value, and the statistic is given by:    

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

z-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".

3) Calculate the statistic    

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

[tex]z=\frac{1165-1150}{\frac{125}{\sqrt{60}}}=0.930[/tex]    

4) Calculate the P-value    

Since is a one-side upper test the p value would be:    

[tex]p_v =P(z>0.930)=1-P(z<0.930)=1-0.824=0.176[/tex]

In Excel we can use the following formula to find the p value "=1-NORM.DIST(0.93,0,1,TRUE)"  

5) Conclusion    

If we compare the p value with a significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the balances of a credit company are significantly higher than $1150 at 0.05 of signficance.    

Answer:

3) x=10

Step-by-step explanation:

9x-16 +3x+11 +7x -5 =180[ sum of

interior angle of

triangle ]

19x -10 =180

19x =190

x=10

4)

2x-12 +4x+43=9x -26

or, 6x + 31= 9x -26

or, 31 +26 = 9x -6x

or, 57 = 3x

or, 19 =x

reason of this question is given up

Answer:

a) 0.8132 or 81.32%

b) 0.1868 or 18.68%

c) 0.6116 or 61.16%

d) 0.7667 minutes

e) 0.8667 minutes

f) 0.1 minutes

Step-by-step explanation:

a)

If f(x) = 0.85-0.35x (0<x<2) is the PDF and X is the random variable that measures the time it takes for a trainee to complete the task, the probability a trainee will complete the task in less than 1.31 minutes is P(X<1.31)

[tex]\large P(X<1.31)=\int_{0}^{1.31}f(x)dx=\int_{0}^{1.31}(0.85-0.35x)dx=\\\\=0.85\int_{0}^{1.31}dx-0.35\int_{0}^{1.31}xdx=\\\\=0.85*1.31-0.35*\frac{(1.31)^2}{2}=0.8132[/tex]

b)

The probability that a trainee will complete the task in more than 1.31 minutes is  

P(X>1.31) = 1 - P(X<1.31) = 1 - 0.8132 = 0.1868

c)

The probability it will take a trainee between 0.25 minutes and 1.31 minutes to complete the task is P(0.25<X<1.31)  

[tex]\large P(0.25<X<1.31)=\int_{0.25}^{1.31}f(x)dx=\int_{0}^{1.31}f(x)dx-\int_{0}^{0.25}f(x)dx=\\\\=0.8132-\int_{0}^{0.25}(0.85-0.35x)dx=0.8132-0.85\int_{0}^{0.25}dx+0.35\int_{0}^{0.25}xdx=\\\\=0.8132-0.85*0.25+0.35*\frac{(0.25)^2}{2}=0.6116[/tex]

d)

the expected time it will take a trainee to complete the task

is E(X)

[tex]\large E(X)=\int_{0}^{2}xf(x)dx=\int_{0}^{2}x(0.85-0.35x)dx=0.85\int_{0}^{2}xdx-0.35\int_{0}^{2}x^2dx=\\\\=0.85*\frac{2^2}{2}-0.35*\frac{2^3}{3}=0.7667\;minutes[/tex]

e)

[tex]\large E(X^2)=\int_{0}^{2}x^2f(x)dx=\int_{0}^{2}x^2(0.85-0.35x)dx=0.85\int_{0}^{2}x^2dx-0.35\int_{0}^{2}x^3dx=\\\\=0.85*\frac{2^3}{3}-0.35*\frac{2^4}{4}=0.8667[/tex]

f)

[tex]\large Var(X)=E(X^2)-(E(X))^2=0.8667-0.7667=0.1000[/tex]

Answer:

The equation of the displacement [tex]d[/tex] as a function of time [tex]t[/tex] is :

[tex]d(t)=8sin(\pi t+\pi )[/tex]

Step-by-step explanation:

Generally , A simple harmonic wave is a sinusoidal function that is it can be expressed in simple [tex]sin[/tex] or [tex]cos[/tex] terms.

Thus,

[tex]d(t) = Asin(wt+c)[/tex]

is the general form of displacement of a SHM.

where,

given,

                               [tex]w=\frac{2\pi} {T}[/tex]

∴

[tex]w[/tex] = [tex]\frac{2\pi }{2}  = \pi[/tex] [tex]rads^{-1}[/tex]

∴

the equation :

⇒[tex]d(t)=8sin(\pi t+c)[/tex]                        ------1

since [tex]d=0[/tex] when [tex]t=o[/tex] ,

⇒[tex]0=8sinc\\c=n\pi[/tex]                         ------2

where n is an integer ;

⇒since the bouy immediately moves in the negative direction , [tex]x[/tex] must be negative or c must be an odd multiple of [tex]\pi[/tex].

⇒ [tex]d(t) = 8sin(\pi t+(2k+1)\pi )[/tex]         ------3

where k is also an integer ;

the least value of [tex]k=0[/tex];

thus ,

the equation is :

[tex]d(t)=8sin(\pi t+\pi )[/tex]

The equation for the displacement d as a function of time t for the buoy is d=-8 cos(πt), as this corresponds to the simple harmonic motion of a system starting at equilibrium and moving downwards.

The motion described by the buoy is that of a simple harmonic oscillator. The displacement (d) at any given time (t), especially for a system initiating motion from the equilibrium and moving downwards, can be given by the equation of a cosine function. Therefore, the equation would be d= - A cos(2πt/T), where A is the amplitude and T is the period. Given the amplitude (A) of 8 in and the period (T) of 2 sec, the equation modeling the displacement d as a function of time t for this buoy can be given by: d = -8 cos(πt).

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Answer: (94.08, 101.92)

Step-by-step explanation:

The confidence interval for unknown population mean[tex](\mu)[/tex] is given by :-

[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

, where [tex]\overline{x}[/tex] = Sample mean

[tex]\sigma[/tex] = Population standard deviation

z* = Critical z-value.

Given : A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.

[tex]\sigma= 12[/tex]

[tex]\overline{x}=98[/tex]

n= 36

Confidence interval = 95%

We know that the critical value for 95% Confidence interval : z*=1.96

Then, the 95% confidence interval for the mean of x  will be :-

[tex]98\pm (1.96)\dfrac{12}{\sqrt{36}}[/tex]

[tex]=98\pm (1.96)\dfrac{12}{6}[/tex]

[tex]=98\pm (1.96)(2)[/tex]

[tex]=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)[/tex]

Hence, the 95% confidence interval for the mean of x is (94.08, 101.92) .

Answer:  95% confidence interval would be (94.08,101.92).

Step-by-step explanation:

Since we have given that

n = 36

standard deviation = 12

sample mean = 98

At 95% confidence, z = 1.96

So, interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=98\pm 1.96\dfrac{12}{\sqrt{36}}\\\\=98\pm 3.92\\\\=(98-3.92,98+3.92)\\\\=(94.08,101.92)[/tex]

Hence, 95% confidence interval would be (94.08,101.92).

Answer:

Which is the output of the formula =AND(12>6;6>3;3>9)?

A.

TRUE

B.

FALSE

C.

12

D.

9

Step-by-step explanation:

Answer:

(a) Yes effect is significant.

(b) 0.15 or 15%

Step-by-step explanation:

The difference in calorie consumption is significantly different from the mean, as indicated by a p value less than 0.05. The proportion of variance, or the size of the effect, is 0.15 or 15%.

For question (a), the effect is indeed significant. This is determined by the 'p' value which is less than 0.05. This shows that the difference between the sample and the mean calories consumed by the average American is statistically significant.

For question (b), the proportion of variance, denoted by η2, is used to measure the size of the effect independent of the sample size. It is the proportion of total variance in the dependent variable that is associated with the membership of different groups formed by the independent variable. η2 in this case is 0.15, which means 15% of the total variation in calorie consumption can be explained by the effect. Hence, the proportion of variance for this effect rounded to two decimal places is 0.15 or 15%.

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Answer: Our required probability is 0.0153.

Step-by-step explanation:

Since we have given that

probability of sample shipped in small packages P(S) = 53% = 0.53

Probability of sample shipped in large packages P(L) = 47% = 0.47

Probability of sample in small break during transit P(S|B)= 2%=0.02

Probability of sample in large break during transit P(L|B) = 1% = 0.01

so, According to bayes theorem, we get that

Proportion of samples break during shipment is given by

[tex]P(S).P(S|B)+P(L).P(L|B)\\\\=0.53\times 0.02+0.47\times 0.01\\\\=0.0153[/tex]

Hence, our required probability is 0.0153.

Final answer:

To find the proportion of samples that break during shipment, multiply the percentage of samples in each packaging type by their breakage rates and add the results. Doing so, we find that 1.47% of the samples break during shipment.

Explanation:

To calculate the proportion of samples that break during shipment, we need to use the percentages of each type of packaging and their respective breakage rates. For small packages, which account for 47% of shipments, a 2% break rate applies. For large packages that make up 53% of the shipments, we have a 1% break rate. The calculation is as follows:

Now we add these two probabilities together to find the total proportion of samples that break:

Total broken samples proportion = (0.47 * 0.02) + (0.53 * 0.01) = 0.0094 + 0.0053 = 0.0147

Thus, the proportion of broken samples during shipment is 0.0147, or 1.47% when rounded to four decimal places.

Answer:

Step-by-step explanation:

See the attached for a graph of 1/sin(x) = csc(x). Just as sin(x) is an odd function, so is 1/sin(x).

Any odd function is symmetrical about the origin, so the left side of the graph is a copy of the right side, rotated 180° about the origin ("in the  other direction"). That is, csc(x) = -csc(-x). This is true because the function is odd, not because it is periodic.

__

Csc(x) is also a periodic function with a period of 2π. That means ...

  csc(x+2π) ≡ csc(x)

This sort of replication of the function is true for all periodic functions (where the added value, the horizontal translation, is the period of the function).

__

You will note that the relation applicable to an odd function (-f(-x) = f(x)) is different from the relation applicable to a periodic function (f(x) = f(x+period)). An odd periodic function, such as csc(x), will be described by both relations.

Answer:

A

Step-by-step explanation:

Answer: number of years that it will take for the balance to reach ​$120,000 is 42 years

Step-by-step explanation:

Initial amount deposited into the account is $4000. This means that the principal is $4000

P = 4000

It was compounded annually. This means that it was compounded once in a year. So

n = 1

The rate at which the principal was compounded is 8.4%. So

r = 8.4/100 = 0.084

Let the number of years that it will take for the balance to reach ​$120,000. It means that it was compounded for a total of t years.

Amount, A at the end of t years is $120,000

The formula for compound interest is

A = P(1+r/n)^nt

120000 = 4000(1 + 0.084/1)^1×t

120000/4000 = 1.084^t

30 = 1.084^t

t = 42 years

Answer:

Step-by-step explanation:

The slope of a perpendicular line is the negative reciprocal of the slope of the original line. Let line x represent the original line and let line y represent the line that is perpendicular to line x

A conditional statement is an if - then statement. It is connected by a hypothesis statement and a conclusion statement. The hypothesis statement is "if the slope of line x is 2/3 and the slope of line y is -3/2 which is its negative reciprocal

The conclusion statement is " then, line x is perpendicular to line y

So the combined statement is

if the slope of line x is 2/3 and the slope of line y is -3/2 which is its negative reciprocal, then line x is perpendicular to line y

Answer:

Step-by-step explanation:

(a) The probability that randomly selected high school student plays organized sport is 236/300 = 59/75 = 0.787 = 78.7%.

(b) The probability means that for every 75 random students selected, 59 of them plays organized sport.

Answer:

Null hypothesis:  [tex]\mu \leq 133[/tex]  

Alternative hypothesis :[tex]\mu>133[/tex]  

Step-by-step explanation:

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".

Data given and notation  

[tex]\bar X[/tex] represent the mean breaking strength value for the sample  

[tex]\sigma=25[/tex] represent the population standard deviation  

[tex]n=100[/tex] sample size  

[tex]\mu_o =133[/tex] represent the value that we want to test  

[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)  

State the null and alternative hypotheses.  

Is a one right tailed test.  

What are H0 and Ha for this study?  

We want to test if the children in this population are gaining weight, so we want to test if the mean increase from the reference value or no.

Null hypothesis:  [tex]\mu \leq 133[/tex]  

Alternative hypothesis :[tex]\mu>133[/tex]  

Final answer:

The correct null and alternative hypotheses for testing whether children are gaining weight compared to a known mean would be H0: μ = 133 and H1: μ > 133, respectively, representing a one-tailed test.

Explanation:

When a student is asking about the null and alternative hypotheses for a population's weight distribution, they are referring to the initial assumptions and contrasting propositions made in a statistical test.

Part (a) of the question suggests that we are trying to test if children in this population are gaining weight. Therefore, the correct set of hypotheses to test this claim would be:
H0: μ = 133 (the null hypothesis, stating that the mean weight is equal to 133 pounds)
H1: μ > 133 (the alternative hypothesis, stating that the mean weight is greater than 133 pounds).

This setup represents a one-tailed test because we are only interested in whether the weight has increased, not if it has changed in any direction.

Answer:

  [tex]f(g(x))=\dfrac{|x|+1-||x|-1|}{2}[/tex]

Step-by-step explanation:

The most straightforward way is direct substitution of g(x) for x in f(x):

  [tex]f(g(x))=\dfrac{|x|+1-||x|-1|}{2}[/tex]

Answer:

a) [tex]P(-1.45<Z<1.45)=0.853[/tex]

b) [tex]P(-1.63<Z<1.63)=0.8968[/tex]

c) [tex]P(-1.48<Z<1.48)=0.8612[/tex]

d) [tex]P(-1.37<Z<1.37)=0.8294[/tex]

Step-by-step explanation:

To find : Determine the area under the standard normal curve that lies ?

Solution :

a) In between Z=-1.45 and Z=1.45

i.e. [tex]P(-1.45<Z<1.45)[/tex]

Now, [tex]P(-1.45<Z<1.45)=P(Z<1.45)-P(Z<-1.45)[/tex]

Using Z-table,

[tex]P(-1.45<Z<1.45)=0.9265-0.0735[/tex]

[tex]P(-1.45<Z<1.45)=0.853[/tex]

b) In between Z=-1.63 and Z=1.63

i.e. [tex]P(-1.63<Z<1.63)[/tex]

Now, [tex]P(-1.63<Z<1.63)=P(Z<1.63)-P(Z<-1.63)[/tex]

Using Z-table,

[tex]P(-1.63<Z<1.63)=0.9484-0.0516[/tex]

[tex]P(-1.63<Z<1.63)=0.8968[/tex]

c) In between Z=-1.48 and Z=1.48

i.e. [tex]P(-1.48<Z<1.48)[/tex]

Now, [tex]P(-1.48<Z<1.48)=P(Z<1.48)-P(Z<-1.48)[/tex]

Using Z-table,

[tex]P(-1.48<Z<1.48)=0.9306-0.0694[/tex]

[tex]P(-1.48<Z<1.48)=0.8612[/tex]

d) In between Z=-1.37 and Z=1.37

i.e. [tex]P(-1.37<Z<1.37)[/tex]

Now, [tex]P(-1.37<Z<1.37)=P(Z<1.37)-P(Z<-1.37)[/tex]

Using Z-table,

[tex]P(-1.37<Z<1.37)=0.9147-0.0853[/tex]

[tex]P(-1.37<Z<1.37)=0.8294[/tex]

Finding the area under the standard normal curve to the left of specific Z-scores involves looking up these Z-values in a Z-table or inputting them into an appropriate calculator. The Z-scores in question range from -1.45 to 1.48 and the related cumulative probabilities represent the area under the curve.

The question relates to finding the area under the standard normal curve, commonly referred to in statistics as Z-scores. These standardized scores indicate how many standard deviations away from the mean a particular point or score is located.

(a) Z equals negative 1.45, meaning it lies 1.45 standard deviations below the mean. When looking up this value in a standard Z-table (or using a calculator), you will find the associated cumulative probability (the area to the left under the curve).

The same process applies to (b) Z equals 0.63 (lies 0.63 standard deviations above the mean), (c) Z equals 1.48 (1.48 standard deviations above the mean) and (d) Z equals negative 1.37 (1.37 standard deviations below the mean). The exact cumulative probabilities vary with each Z-score and these represent the area under the curve to the left of each point.

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Answer:

The instantaneous velocity for [tex]s(t) = 8t + 19[/tex] when t = 4 is [tex]v(t)=8 \:\frac{ft}{s}[/tex].

Step-by-step explanation:

The average rate of change of function f over the interval is given by this expression:

[tex]\frac{f(b)-f(a)}{b-a}[/tex]

The average velocity is the average rate of change of distance with respect to time

[tex]average \:velocity=\frac{distance \:traveled}{time \:elapsed} =\frac{\Delta s}{\Delta t}[/tex]

The instantaneous rate of change is defined to be the result of computing the average rate of change over smaller and smaller intervals.

The derivative of f with respect to x, is the instantaneous rate of change of f with respect to x and is thus given by the formula

[tex]f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]

For any equation of motion s(t), we define what we call the instantaneous velocity at time t to be the limit of the average velocity, between t and t + Δt, as Δt approaches 0.

[tex]v(t)=\lim_{\Delta t \to 0} \frac{s(t+\Delta t)-s(t)}{\Delta t}[/tex]

We know the equation of motion [tex]s(t) = 8t + 19[/tex] and we want to find the instantaneous velocity for the given value of t = 4.

Applying the definition of instantaneous velocity, we get

[tex]v(t)=\lim_{\Delta t \to 0} \frac{s(t+\Delta t)-s(t)}{\Delta t}\\\\v(4)=\lim_{\Delta t \to 0} \frac{8(4+\Delta t)+19-(8(4)+19)}{\Delta t}\\\\v(t)=\lim_{\Delta t \to 0} \frac{32+8\Delta t+19-32-19}{\Delta t}\\\\v(t)=\lim_{\Delta t \to 0} \frac{8\Delta t}{\Delta t}\\\\v(t)=8 \:\frac{ft}{s}[/tex]

The instantaneous velocity for [tex]s(t) = 8t + 19[/tex] when t = 4 is [tex]v(t)=8 \:\frac{ft}{s}[/tex].

The volume of a box is the amount of space in the box.

(a) The dimension of the box

Let:

The width of the cardboard be x.

So, the length of the cardboard is: 2.4x

When 3 inches is removed, the dimension of the box is:

[tex]\mathbf{L = 2.4x- 6}[/tex] --- length

[tex]\mathbf{W = x - 6}[/tex] ---- width

[tex]\mathbf{H = 3}[/tex] --- height

(b) The restriction on x

When 3 inches is removed, it means that a total of 6 inches will be removed from either sides.

Hence, the value of x must be greater than 6.

So, the restriction on x is:

[tex]\mathbf{x > 6}[/tex]

(c) Function that represents volume.

The volume (V) of a box is:

[tex]\mathbf{V = L \times W \times H}[/tex]

So, we have:

[tex]\mathbf{V = (2.4x - 6) \times (x - 6) \times 3}[/tex]

Simplify

[tex]\mathbf{V = (7.2x - 18) \times (x - 6)}[/tex]

Expand

[tex]\mathbf{V = 7.2x^2 - 43.2x - 18x +108}[/tex]

[tex]\mathbf{V = 7.2x^2 -61.2x +108}[/tex]

(d) The dimension, when the volume is 520

This means that, V = 520

So, we have:

[tex]\mathbf{ 7.2x^2 -61.2x +108 = 520}[/tex]

Collect like terms

[tex]\mathbf{ 7.2x^2 -61.2x +108 - 520 = 0}[/tex]

[tex]\mathbf{ 7.2x^2 -61.2x -412 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{x = 12.93\ or\ x = -4.43}[/tex]

Recall that: [tex]\mathbf{x > 6}[/tex]

So, the value of x is:

[tex]\mathbf{x = 12.93}[/tex]

Substitute 12.93 for x in [tex]\mathbf{L = 2.4x- 6}[/tex]  and [tex]\mathbf{W = x - 6}[/tex]

So, we have:

[tex]\mathbf{L = 2.4 \times 12.93 - 6 = 25.03}[/tex]

[tex]\mathbf{W = 12.93 - 6 = 6.93}[/tex]

So, the dimension of the box is: 25.03 by 6.93 by 3

(e) The value of x, when the volume is between 600 and 800

This means that, V = 600 and V = 800

When V = 600, we have:

[tex]\mathbf{ 7.2x^2 -61.2x +108 = 600}[/tex]

Collect like terms

[tex]\mathbf{ 7.2x^2 -61.2x +108 - 600 = 0}[/tex]

[tex]\mathbf{ 7.2x^2 -61.2x -492 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{x = 13.54\ or\ x = -5.04}[/tex]

Recall that: [tex]\mathbf{x > 6}[/tex]

So, the value of x is:

[tex]\mathbf{x = 13.54}[/tex]

When V = 800, we have:

[tex]\mathbf{ 7.2x^2 -61.2x +108 = 800}[/tex]

Collect like terms

[tex]\mathbf{ 7.2x^2 -61.2x +108 - 800 = 0}[/tex]

[tex]\mathbf{ 7.2x^2 -61.2x -692 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{x = 14.93\ or\ x = -6.45}[/tex]

Recall that: [tex]\mathbf{x > 6}[/tex]

So, the value of x is:

[tex]\mathbf{x = 14.93}[/tex]

If such a box is to have a volume between 600 and 800, the value of x would be between 13.54 and 14.93

Read more about volumes at:

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The response provides step-by-step explanations and formulas for finding the length, restrictions, volume, and dimensions of a box made from a cardboard piece. It also gives the dimensions that correspond to a specific volume and the range of values that produce volumes within a given range.

a) The length of the original piece of cardboard can be represented as 2.4 times the width, so it is 2.4x inches long.

b) The restrictions on x are x>6. The dimensions of the bottom rectangular base of the box are (2.4x-6) inches for the length and (x-6) inches for the width.

c) The function V that represents the volume of the box in terms of x is V = (2.4x-6)(x-6)(x-6).

d) To find the dimensions of the bottom of the box when the volume is 520 in^3, we can solve the equation V = 520. The length will be approximately 25.0 inches and the width will be approximately 6.9 inches.

e) To find the values of x for a volume between 600 and 800 in^3, we can solve the inequalities 600 <= V <= 800. The range of values for x is approximately 11.9 to 16.9 inches.

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Answer:

with

what maam

Step-by-step explanation:

Answer: the correct option is D

Step-by-step explanation:

The zoologist is recording the loss of wolves in her state and she recorded that the number of wolves, w, in the state on January 1.

One year later, there were 84 wolves in the state which is 23 fewer than the number of wolves in the state a year earlier. This means that a year later, the number of wolves was 23 fewer than w

The number of wolves in the state on January 1 would be the sum of the number of wolves in the state a year later and the number by which it became fewer than the number on January 1. This becomes

w = 84 + 23

w - 23 = 84

Answer:  0.0918, it is not unusual.

Step-by-step explanation:

Given : The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.54 minutes and a standard deviation of 1.91.

i.e. [tex]\mu=8.54[/tex] minutes and [tex]\sigma= 1.91[/tex] minutes

Let x denotes the length of time a person takes to decide which shoes to purchase.

Formula : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]

Then, the probability that a randomly selected individual will take less than 6 minutes to select a shoe purchase will be :-

[tex]\text{P-value=}P(x<6)=P(\dfrac{x-\mu}{\sigma}<\dfrac{6-8.54}{1.91})\\\\\approx P(z<1.33)=1-P(z<1.33)\ \ \ [\becaus\ P(Z<-z)=1-P(Z<z)]\\\\=1-0.9082\ \ [\text{By using z-value}]=0.0918[/tex]

Thus , the required probability = 0.0918

Since, P-value (0.0918) >0.05 , it means this outcome is not unusual.

[Note : When a outcome is unusual then the probability of its happening is less than or equal to 0.05. ]

11/18 is greater than ( > ) 5/9

Answer:

relative minimum

Step-by-step explanation:

From f^''(x) = 12x we can take the integration to find out what f'(x) is:

[tex]f'(x) = 6x^2 + C[/tex]

Furthermore, we can substitute x = 1 for f''(x) to find out whether it's positive or negative

f''(1) = 12*1 = 12 > 0

So if x=1 is a critical point of f'(x) and f''(x=1) > 0 then that point is a relative minimum point  

Using the second derivative test with the given information f"(x)=12x, it's concluded that f(x) has a relative minimum at x=1, as f"(1) is positive.

To determine if f(x) has a relative minimum or maximum at x=1, we can use the given information that f"(x)=12x and that x=1 is a critical number of f'(x). Since the second derivative, f"(x), is 12x, at x = 1, it is positive (f"(1)=12(1)=12). According to the second derivative test, a positive second derivative at a critical point indicates that the function has a relative minimum at that point. Therefore, we can conclude that f(x) has a relative minimum at x=1.

Answer:

See steps below

Step-by-step explanation:

The function  

[tex]\large f(x,y)=(1-x)^2+100(y-x^2)^2[/tex]

is a particular case of the general Rosenbrock’s Function.

a)  

Since  

[tex]\large ((1-x)^2\geq 0[/tex] for all the values of x and equals 0 when x=1 and

[tex]\large (y-x^2)^2=(y-1)^2\geq 0[/tex] for all the values of y and equals 0 only when y=1, we conclude that (1,1) is a minimum.

On the other hand,

f(x,y)>0 for (x,y) ≠ (1,1) so (1,1) is a global minimum.

b)

To confirm that this function is not convex, we will be using the following characterization of convexity

“f is convex if, and only if, the Laplace operator of f [tex]\large \nabla^2f \geq 0[/tex] for every (x,y) in the domain of f”

Given that the domain of f is the whole plane XY, in order to prove that f is not convex, we must find a point (x,y) at where the Laplace operator is < 0.

The Laplace operator is given by

[tex]\large \nabla^2f=\displaystyle\frac{\partial ^2f}{\partial x^2}+\displaystyle\frac{\partial ^2f}{\partial y^2}[/tex]

Let us compute the partial derivatives

[tex]\large \displaystyle\frac{\partial f}{\partial x}=-2(1-x)+200(y-x^2)(-2x)=-2+2x-400xy+400x^3\\\\\displaystyle\frac{\partial^2 f}{\partial x^2}=2-400y+1200x^2[/tex]

and

[tex]\large \displaystyle\frac{\partial f}{\partial y}=200(y-x^2)=200y-200x^2\\\\\displaystyle\frac{\partial^2 f}{\partial x^2}=200[/tex]

we have then

[tex]\large \nabla^2 f=2-400y+1200x^2+200[/tex]

if we take (x,y) = (0,1)

[tex]\large \nabla^2 f(0,1)=2-400+200=202-400=-198<0[/tex]

hence f is not convex.

Answer:

[tex]\frac{7}{6} hours[/tex]

Step-by-step explanation:

Distance for one way = 14 miles

The boat goes an average speed of 20 mph

The train goes 10 mph faster than the boat.

Speed of train = 20+10 = 30 kmph

Abdul takes the train into the park and the boat on his way back.

Time taken for going into park by train = [tex]\frac{Distance}{\text{Speed of train}}[/tex]

                                                                 = [tex]\frac{14}{30}[/tex]

Time taken for going back by boat = [tex]\frac{Distance}{\text{Speed of train}}[/tex]

                                                                 = [tex]\frac{14}{20}[/tex]

So, total time =  [tex]\frac{14}{30}+\frac{14}{20}=\frac{7}{6} hours[/tex]

Hence the round trip​ take [tex]\frac{7}{6} hours[/tex]

The period of the round trip is mathematically given as

T=7/6hours

Question Parameter(s):

guests have to take either a train or a boat 14 miles

The train goes 10 mph faster than the boat.

The boat goes an average speed of 20 mph.

Generally, the equation for the time   is mathematically given as

t=d/v

Therefore

For going fort

t=14/30

For going back

t'= 14/20

In conclusion, total time

[tex]T= \frac{14}{30}+\frac{14}{20}[/tex]

T=7/6

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Answer:

Step-by-step explanation:

Answer:

The particles collide when t = 7 at the point (49, 49, 49).

Step-by-step explanation:

We know the trajectories of the two particles,

[tex]r_1(t)=\langle t^2,16t-63,t^2\rangle\\r_2(t)=\langle 9t-14,t^2,13t-42\rangle[/tex]

To find if the tow particles collide you must:

[tex]t^2=9t-14\\t^2-9t+14=0\\\left(t^2-2t\right)+\left(-7t+14\right)=0\\t\left(t-2\right)-7\left(t-2\right)=0\\\left(t-2\right)\left(t-7\right)=0[/tex]

The solutions to the quadratic equation are:

[tex]t=2,\:t=7[/tex]

[tex]16t-63=t^2\\^2-16t+63=0\\\left(t^2-7t\right)+\left(-9t+63\right)=0\\t\left(t-7\right)-9\left(t-7\right)=0\\\left(t-7\right)\left(t-9\right)=0[/tex]

The solutions to the quadratic equation are:

[tex]t=7,\:t=9[/tex]

[tex]t^2=13t-42\\t^2-13t+42=0\\\left(t^2-6t\right)+\left(-7t+42\right)=0\\t\left(t-6\right)-7\left(t-6\right)=0\\\left(t-6\right)\left(t-7\right)=0[/tex]

The solutions to the quadratic equation are:

[tex]t=6,\:t=7[/tex]

Evaluate the position vectors at the common time. The common solution is when t = 7.

[tex]r_1(7)=\langle 7^2,16(7)-63,7^2\rangle=\langle 49,49,49\rangle\\\\r_2(7)=\langle 9(7)-14,7^2,13(7)-42\rangle=\langle 49,49,49\rangle[/tex]

For two particles to collide, they must be at exactly the same coordinates at exactly the same time.

The particles collide when t = 7 at the point (49, 49, 49).

(a) Hypotheses:

[tex]\[ H_0: \mu = 3 \]\\H_a: \mu \neq 3 \][/tex]

(b) **Test Statistic and P-Value:

[tex]\[ t \approx 1.878 \][/tex]

(c) [tex]\[ \text{P-Value} \approx 0.0678 \][/tex]

Let's perform a hypothesis test to determine whether the number of weeks until a house is sold in Greene County differs from the national average. The average number of weeks to sell a home nationally is 3 weeks.

Hypotheses:

- Null hypothesis [tex](\(H_0\))[/tex]: The average number of weeks to sell a home in Greene County is equal to the national average [tex](\(μ = 3\) weeks)[/tex].

[tex]\[ H_0: \mu = 3 \][/tex]

- Alternative hypothesis [tex](\(H_a\))[/tex]: The average number of weeks to sell a home in Greene County differs from the national average.

[tex]\[ H_a: \mu \neq 3 \][/tex]

Given Information:

- Sample mean [tex](\(\bar{x}\))[/tex] = 3.6 weeks

- Sample standard deviation s = 2 weeks

- Sample size n = 39

- Level of significance [tex](\(\alpha\))[/tex] = 0.05

Test Statistic:

The test statistic for a one-sample t-test is given by:

[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \][/tex]

Calculation:

[tex]\[ t = \frac{3.6 - 3}{\frac{2}{\sqrt{39}}} \][/tex]

Let's calculate the value of \(t\) and the p-value.

Calculation:

[tex]t = \frac{3.6 - 3}{\frac{2}{\sqrt{39}}} \]\\t \approx \frac{0.6}{\frac{2}{\sqrt{39}}} \]\\t \approx \frac{0.6}{0.319439} \]\\t \approx 1.878 \][/tex]

Degrees of Freedom:

The degrees of freedom for a one-sample t-test is \(n-1\), where \(n\) is the sample size.

[tex]\[ \text{Degrees of Freedom} = 39 - 1 = 38 \][/tex]

P-Value:

Now, we need to find the p-value associated with this t-statistic and degrees of freedom. Since it's a two-tailed test, we're interested in the probability that the t-statistic is greater than 1.878 or less than -1.878.

You can use a t-table or statistical software to find this p-value. For [tex]\(df = 38\)[/tex] and [tex]\(t \approx 1.878\)[/tex], the p-value is approximately 0.0678 (two-tailed).

Conclusion:

Since the p-value (0.0678) is greater than the significance level (\(\alpha = 0.05\)), we do not reject the null hypothesis.

(a) Hypotheses:

[tex]\[ H_0: \mu = 3 \]\\H_a: \mu \neq 3 \][/tex]

(b) Test Statistic and P-Value:

[tex]\[ t \approx 1.878 \][/tex]

[tex]\[ \text{P-Value} \approx 0.0678 \][/tex]

(c) Conclusion:

At a 5% level of significance, there is not enough evidence to conclude that the average number of weeks until a house is sold in Greene County differs from the national average in 2017.

(a)State the null and alternative hypothesis. (Enter != for ≠ as needed.)H0:H0: μ = 3 (weeks) Ha: μ ≠ 3 (weeks)(b) the value of the test statistic. (Round your answer to three decimal places.) the p-value. (Round your answer to four decimal places.)p-value =1.702

To conduct the hypothesis test, calculate the test statistic and the p-value. The test statistic for a hypothesis test comparing a sample mean to a population mean when the population standard deviation is known is calculated using the formula:

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

Where:

[tex]\(\bar{X}\)[/tex]n (3.6 weeks)

\(\mu\) = population mean (3 weeks)

[tex]\(\sigma\)[/tex]n standard deviation (2 weeks)

[tex]\(n\)[/tex]mple size (39)

Plugging in the values:

[tex]\[ Z = \frac{3.6 - 3}{\frac{2}{\sqrt{39}}} \][/tex]

[tex]\[ Z ≈ 1.702 \][/tex]

The p-value associated with this test statistic can be found using a standard normal distribution table or a statistical calculator. With a Z-score of 1.702, the corresponding two-tailed p-value is approximately 0.0883.

The calculated test statistic, \( Z ≈ 1.702 \), falls within the critical region for a two-tailed test at a significance level of α = 0.05. This means that the p-value (0.0883) is greater than the chosen level of significance (0.05). Consequently, there is insufficient evidence to reject the null hypothesis.

Thus, based on the sample data from Greene County, Ohio, in 2017, we do not have enough evidence to conclude that the average number of weeks until a house sold differs significantly from the national average in 2017. This suggests that, statistically, the time it takes for a house to sell in Greene County during that year might not be significantly different from the national average.

Overall, with a p-value of 0.0883 and at a significance level of 0.05, the analysis does not provide enough evidence to reject the null hypothesis, suggesting that the average number of weeks until a house sold in Greene County, Ohio, in 2017 may not differ significantly from the national average in the same year.

Answer:

The 95% confidence interval to estimate is 475.4592 to 484.5408

Step-by-step explanation:

Consider the provided information.

25 pieces of this metallic glass were randomly sampled from a recent production run.

That means the value of n is 25.

The degree of freedom is:

df = n-1

df = 25-1 = 24

X = 480°F and s = 11°F

We need to Construct a 95% confidence interval to estimate.

1-α=0.95

α=0.05

It is a two tail test with small sample size.

Determine the value of t by using Degrees of freedom and Significance level:

The required t value is 2.064

[tex]95\% CI=\bar x\pm t_c\times \frac{s}{\sqrt{n}}[/tex]

Substitute the respective values as shown:

[tex]95\% CI=480\pm 2.064\times \frac{11}{\sqrt{25}}[/tex]

[tex]95\% CI=480\pm 4.5408[/tex]

[tex]95\% CI=475.4592\ to\ 484.5408[/tex]

Hence, the 95% confidence interval to estimate is 475.4592 to 484.5408

Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval to estimate the temperature at which brittleness first appeared is given by (475.46 ºF, 484.54 ºF).

The confidence interval is:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

In which:

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 25 - 1 = 24 df, is t = 2.0639.

The other parameters are as follows:

[tex]n = 25, \overline{x} = 480, s = 11[/tex]

Hence:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 480 - 2.0639\frac{11}{\sqrt{25}} = 475.46[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 480 + 2.0639\frac{11}{\sqrt{25}} = 484.54[/tex]

The 95% confidence interval is (475.46 ºF, 484.54 ºF).

More can be learned about the t-distribution at https://brainly.com/question/16162795