Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 10pm and 3 am, find the probability that:
a) Exactly two will be drunken drivers.
b) Three or four will be drunken drivers.
c) At least 7 will be drunken drivers.
d) At most 5 will be drunken drivers.

Answer:

Toy truck. 1 lb is equal to 453 grams so a toy truck would be less than a lb

Answer:

Step-by-step explanation:

was

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:

(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:

A

Step-by-step explanation:

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.

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

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:

a=55

b=5

T=17

Step-by-step explanation:

The general form of the equation is:

[tex]N(t)=ab^{\frac{t}{T}}[/tex]

For t = 0:

[tex]N(0)=ab^{\frac{0}{T}}\\N(0) = a = 55\\a=55[/tex]

Since there has been a fivefold increase after 17 years, at t = 17, N(17) = 55*5

[tex]N(17)=55b^{\frac{17}{T}}\\55*5 = 55b^{\frac{17}{T}}\\b^{\frac{17}{T}} = 5[/tex]

If at every 17*n years there in an increase of 5^n, one can deduct that the values for T and b are respectively 17 and 5:

[tex]b^{\frac{t}{T}}= 5^{\frac{17n}{17}}[/tex]

Therefore, the function that represents N(t) is:

[tex]N(t)=55*5^{\frac{t}{17}}[/tex]

The constants for the function representing the number of households with cable TV service are: a = 55 (initial number of households), b = 5 (fivefold increase), and T = 17 (the number of years for each fivefold increase). The function is N = 55 x 5^(t/17).

The student is asking to define the constants a, b, and T for the function representing the number of households N with cable TV service. The function is of the form N = abt/T, where t is the time in years since the households were first measured, N initially is 55, and the number of households increases fivefold every 17 years.

To find constants a, b, and T, first note that a is the initial value of households with cable TV service, so a = 55. Since the number increases fivefold every 17 years, b is the factor of increase which is 5. The time period for this increase, T, is every 17 years, so T = 17.

Therefore, the function representing the number of households with cable TV service is N = 55 x 5t/17.

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:

Mass

[tex]\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)[/tex]

Center of mass

Coordinate x

[tex]\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

Coordinate y

[tex]\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

Coordinate z

[tex]\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

Step-by-step explanation:

Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions

x(t) = t

y(t) = 4 cos(t)

z(t) = 4 sin(t)  

for 0 ≤ t ≤ 2π

If D(x,y,z) is the density of W at a given point (x,y,z), the mass  m would be the curve integral along the path W

[tex]m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt[/tex]

The density D(x,y,z) is given by

[tex]D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16[/tex]

on the other hand

[tex]||W'(t)||=\sqrt{1^2+(-4sin(t))^2+(4cos(t))^2}=\sqrt{1+16}=\sqrt{17}[/tex]

and we have

[tex]m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt=\\\\\sqrt{17}\displaystyle\int_{0}^{2\pi}(t^2+16)dt=\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)[/tex]

The center of mass is the point [tex](\bar x,\bar y,\bar z)[/tex]

where

[tex]\bar x=\displaystyle\frac{1}{m}\displaystyle\int_{W}xD(x,y,z)\\\\\bar y=\displaystyle\frac{1}{m}\displaystyle\int_{W}yD(x,y,z)\\\\\bar z=\displaystyle\frac{1}{m}\displaystyle\int_{W}zD(x,y,z)[/tex]

We have

[tex]\displaystyle\int_{W}xD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}t(t^2+16)dt=\\\\=\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)[/tex]

so

[tex]\bar x=\displaystyle\frac{\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

[tex]\displaystyle\int_{W}yD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}4cos(t)(t^2+16)dt=\\\\=16\sqrt{17}\pi[/tex]

[tex]\bar y=\displaystyle\frac{16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

[tex]\displaystyle\int_{W}zD(x,y,z)=4\sqrt{17}\displaystyle\int_{0}^{2\pi}sin(t)(t^2+16)dt=\\\\=-16\sqrt{17}\pi[/tex]

[tex]\bar z=\displaystyle\frac{-16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

Final answer:

To find the mass and center of mass of the wire loop in the shape of a helix, we need to integrate the density function along the length of the wire. The formula for the mass of a wire is m = ∫(ρ dl), where ρ is the density function and dl is an infinitesimally small segment of the wire. To find the center of mass, we can use the formulas Cx = (1/m) ∫(x ρ dl), Cy = (1/m) ∫(y ρ dl), and Cz = (1/m) ∫(z ρ dl), where x, y, and z are the parametric equations and ρ is the density function.

Explanation:

To find the mass and center of mass of the wire loop in the shape of a helix, we need to integrate the density function along the length of the wire. Since the density is equal to the square of the distance from the origin to the point, we can use the formula for the mass of a wire:

m = ∫(ρ dl)

Where ρ is the density function and dl is an infinitesimally small segment of the wire. In this case, we have:

ρ = r² = (x² + y² + z²)

Using the given parametric equations for x, y, and z, we can substitute them into the formula and integrate along the range of t:

m = ∫(t² + (4cos(t))² + (4sin(t))²) dt

Next, to find the center of mass, we can use the formula:

Cx = (1/m) ∫(x ρ dl)

Cy = (1/m) ∫(y ρ dl)

Cz = (1/m) ∫(z ρ dl)

We can substitute the parametric equations for x, y, and z, and the density function into these formulas and integrate along the range of t to find the center of mass.

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:

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:

z=1.461, fail to reject the null hypothesis since [tex]p_v>\alpha[/tex] at 5% of singificance.

Step-by-step explanation:

1) Data given and notation

n=53 represent the random sample taken

X=41 represent the adults with damage liability claims by single people under the age of 25.

[tex]\hat p=\frac{41}{53}=0.774[/tex] estimated proportion of adults with damage liability claims by single people under the age of 25.

[tex]p_o=0.68[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company (68%).:  

Null hypothesis:[tex]p\leq 0.68[/tex]  

Alternative hypothesis:[tex]p > 0.68[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.774 -0.68}{\sqrt{\frac{0.68(1-0.68)}{53}}}=1.461[/tex]

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a unilateral right tailed test the p value would be:  

[tex]p_v =P(z>1.461)=0.072[/tex]  

So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with damage liability claims by single people under the age of 25 is not significantly higher from 0.68 or 68% .  

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. ]

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

Answer:

Step-by-step explanation:

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

Read more about Time

https://brainly.com/question/4931057

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:

a) [tex]\nabla F=0, \nabla\times F=(0,0,-2)[/tex]

b)  [tex]\nabla F=0, \nabla\times F=(0,0,0)[/tex], [tex]f=2xy+3yz+C, C\in\mathbb{R}[/tex].

c)  [tex]\nabla F=4x, \nabla\times F=(0,-4z,0)[/tex]

Step-by-step explanation:

Remember, if F= <f,g,h> is a vector field and [tex]\nabla[/tex] is the operator [tex]\nabla=<\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}>[/tex]

a) [tex]F=(x+y,-x+y,-2z)[/tex]

The divergence of F is

[tex]\nabla F=\frac{\partial}{\partial x}(x+y)+\frac{\partial}{\partial y}(-x+y)+\frac{\partial}{\partial z} (-2z)=1+1-2=0[/tex]

The curl of F is

[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x+y&-x+y&-2z\end{array}\right])\\=i(\frac{\partial}{\partial y}(-2z)-\frac{\partial}{\partial z}(-x+y))-j(\frac{\partial}{\partial x}(-2z)-\frac{\partial}{\partial z}(x+y))+k(\frac{\partial}{\partial x}(-x+y)-\frac{\partial}{\partial y}(x+y))=0i-0j-2k=(0,0,-2)[/tex]

b) [tex]F=(2y,2x+3z,3y)[/tex]

The divergence of F is

[tex]\nabla F=\frac{\partial}{\partial x}(2y)+\frac{\partial}{\partial y}(2x+3z)+\frac{\partial}{\partial z} (3y)=0+0+0=0[/tex]

The curl of F is

[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\2y&2x+3z&3y\end{array}\right])\\=i(\frac{\partial}{\partial y}(3y)-\frac{\partial}{\partial z}(2x+3z))-j(\frac{\partial}{\partial x}(3y)-\frac{\partial}{\partial z}(2y))+k(\frac{\partial}{\partial x}(2x+-3z)-\frac{\partial}{\partial y}(2y))=0i-0j+0k=(0,0,0)[/tex]

Since the curl of F is 0 the we will try find f such that the gradient of f be F.

Since [tex]f_x=2y[/tex], [tex]f=2xy+g(y,z)[/tex].

Since [tex]f_y=2x+3z[/tex], [tex]2x+3z=f_y=2x+g_y(y,z)\\3z=g_y[/tex].

Since [tex]f_z=3y[/tex] and [tex]f_z=3y+h'(z)[/tex], then [tex]h'(z)=0[/tex]. Thins means that [tex]h(z)=C, C\in\mathbb{R}[/tex]

Therefore,

[tex]f=2xy+3yz+C, C\in\mathbb{R}[/tex].

c) [tex]H=(x^2-z^2,2,2xz)[/tex]

The divergence of F is

[tex]\nabla F=\frac{\partial}{\partial x}(x^2-z^2)+\frac{\partial}{\partial y}(2)+\frac{\partial}{\partial z} (2xz)=2x+0+2x=4x[/tex]

The curl of F is

[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x^2-z^2&2&2xz\end{array}\right])\\=i(\frac{\partial}{\partial y}(2xz)-\frac{\partial}{\partial z}(2))-j(\frac{\partial}{\partial x}(2xz)-\frac{\partial}{\partial z}(x^2-z^2))+k(\frac{\partial}{\partial x}(2)-\frac{\partial}{\partial y}(x^2-z^2))=0i-(2z-(-2z))j-0k=(0,-4z,0)[/tex]

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.    

Final answer:

To assess if the sample of fish with high mercury levels supports the hypothesis of a 24% proportion in the population, we set up null and alternative hypotheses, calculate the z-test statistic, compare it to a critical value, compute the p-value, and then make a conclusion based on whether the p-value is less than the chosen significance level of 0.05.

Explanation:

To test the hypothesis about the percentage of freshwater fish with high levels of mercury, we must first establish our null hypothesis (H0) and alternative hypothesis (Ha). The null hypothesis is that the true proportion of fish with dangerous mercury levels is 24% (H0: p = 0.24), while the alternative hypothesis is that the proportion is not 24% (Ha: p ≠ 0.24).

We then calculate the z-test statistic using the sample proportion (p' = 80/350 = 0.2286) and the assumed population proportion under H0 (p = 0.24), along with the standard deviation of the sampling distribution (σp = sqrt[p(1-p)/n]). This result is compared to the critical z-score of 1.96, which corresponds to our significance level (α) of 0.05 in a two-tailed test.

The p-value is calculated to determine the probability of observing our sample statistic, or one more extreme if the null hypothesis is true. If this p-value is less than 0.05, we reject H0 and conclude that the sample provides enough evidence against the null hypothesis.

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:

It will leave the sprinkler at speed of [tex]v_2=613.87m/sec[/tex]

Step-by-step explanation:

We have given internal diameter of the garden hose [tex]d_1=0.740in=1.8796cm[/tex]

So radius [tex]r_1=\frac{d_1}{2}=\frac{1.8796}{2}=0.9398cm[/tex]

So area [tex]A_1=\pi r_1^2=3.14\times 0.9398^2=2.7733cm^2[/tex]

Water in the hose has a speed of 4 ft/sec

So [tex]v_1=4ft/sec=121.92cm/sec(As\ 1ft/sec\ =30.48cm/sec)[/tex]

Number of holes n = 36

Diameter of each hole [tex]d_2=0.1397cm[/tex]

So radius [tex]r_2=0.0698cm[/tex]

So area [tex]A_2=\pi r^2=3.14\times 0.0698^2=0.0153cm^2[/tex]

From continuity equation

[tex]A_1v_1=nA_2v_2[/tex]

[tex]2.7733\times 121.92=36\times 0.0153\times v_2[/tex]

[tex]v_2=613.87m/sec[/tex]

Answer:

  a) 1.30656∠(3π/8), 0.541196∠(15π/8)

  b) 1.30656∠(π/8), 0.541196∠(5π/8)

Step-by-step explanation:

The critical points can be found in polar coordinates by considering ...

  [tex]\displaystyle{dy\over dx}={dy/d\theta\over dx/d\theta}={r\cos\theta + r'\sin\theta\over -r\sin\theta + r'\cos\theta} \quad\text{where $r'=dr/d\theta$}[/tex]

We can simplify the effort a little bit by rewriting r as …

  [tex]r=\sqrt{2}\sin{(\theta+\pi /4)}[/tex]

Then, filling in function and derivative values, we have …

  [tex]\dfrac{dy}{dx}=\dfrac{\sqrt{2}(\sin{(\theta+\pi /4)}\cos{(\theta)}+\cos{(\theta+\pi /4)}\sin{(\theta)})}{\sqrt{2}(-\sin{(\theta+\pi /4)}\sin{(\theta)}+\cos{(\theta+\pi /4)}\cos{(\theta)})}\\\\=\dfrac{\sin{(2\theta+\pi /4)}}{\cos{(2\theta+\pi /4)}}\\\\\dfrac{dy}{dx}=\tan{(2\theta +\pi /4)}[/tex]

__

(a) For horizontal tangents, dy/dx = 0, so we have …

 [tex]\tan{(2\theta+\pi /4)}=0\\\\2\theta+\dfrac{\pi}{4}=k\pi \quad\text{for some integer k}\\\\\theta=k\dfrac{\pi}{2}-\dfrac{\pi}{8}[/tex]

We can use reference angles for the “r” expressions and write the two horizontal tangent point (r, θ) values of interest as …

  [tex](\sqrt{2}\sin{\dfrac{3\pi}{8}},\dfrac{3\pi}{8})\ \text{and}\ (\sqrt{2}\sin{\dfrac{\pi}{8}},\dfrac{15\pi}{8})[/tex]

__

(b) For vertical tangents, dy/dx = undefined, so we have …

 [tex]2\theta+\dfrac{\pi}{4}=k\pi +\dfrac{\pi}{2} \quad\text{for some integer k}\\\\\theta=k\dfrac{\pi}{2}+\dfrac{\pi}{8}[/tex]

Again using reference angles for “r”, the two vertical tangent point values of interest are …

  [tex](\sqrt{2}\sin{\dfrac{3\pi}{8}},\dfrac{\pi}{8})\ \text{and}\ (\sqrt{2}\sin{\dfrac{\pi}{8}},\dfrac{5\pi}{8})[/tex]

__

The attached graph shows the angle values in degrees and the radius values as numbers. The points of tangency are mirror images of each other across the line y=x.

To find points with horizontal tangents for the polar curve r = cos(θ) + sin(θ), one must set the derivative with respect to θ to zero; for vertical tangents, find where the derivative is undefined. The angles found are then substituted back into the equation to determine r values, and points are listed within the specified domain of θ and r.

The question deals with finding points where the tangent lines to the polar curve r = cos(θ) + sin(θ) are horizontal or vertical. To find where the tangent lines are horizontal, we need to look for points where the derivative of the function with respect to θ is zero, since a horizontal tangent line would have a slope of zero. Conversely, for vertical tangent lines, we want to find where the derivative is undefined or infinite.

A general method for finding this for a polar curve involves taking the derivative of r with respect to θ and setting it equal to zero for horizontal tangents, and finding where it is undefined for vertical tangents:

Horizontal tangent: [tex]\frac{dr}{dθ}[/tex]= 0

Vertical tangent:  [tex]\frac{dr}{dθ}[/tex] is undefined

Once the relevant angles are determined, we substitute them back into the original equation to find the corresponding r value, ensuring that we only consider points within the given domain of θ from 0 to 2π and r ≥ 0.

For the given polar curve r = cos(θ) + sin(θ), we take the derivative of r with respect to θ, which gives us -sin(θ) + cos(θ). Setting this equal to zero yields angles where horizontal tangents occur. Points where this derivative is undefined will give us the angles for vertical tangents. The corresponding points (r, θ) are then listed according to the instructions.

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:

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:

Greatest area: One square enclosure with side 50 m

Least area: Two square enclosures with side 25 m each

Step-by-step explanation:

We know we have 200 m of fencing to make the required enclosures. Since the fence will run surrounding any enclosure, it is called the perimeter.

The perimeter for any square area of side z is computed as

P=4z

And its area is

[tex]A=z^2[/tex]

Now, let's analyze both options (shown in the figure below)

Option 1: One square enclosure

Knowing P=200 m, we can determine the length of the side

z=200 m /4 = 50 m

The area is easily computed

[tex]A=50^2=2500 m^2[/tex]

Option 2: Two separate square enclosures of any size

Let's say the side of one of them is x and the side of the other one is y

Assuming both enclosures have no sides in common, the total perimeter is

P=4x+4y

We have 200 m to make the job, so

4x+4y=200

Or equivalently

x+y=50 => y=50-x

The total area of both enclosures is

[tex]A=x^2+y^2[/tex]

Replacing the expression of y

[tex]A=x^2+(50-x)^2[/tex]

To know what the best value is for x to maximize or minimize the area, we use derivatives with respect to x

[tex]A'=2x+2(50-x)(-1)[/tex]

[tex]A'=2x-100+2x=4x-100[/tex]

We equate A'=0 to find the critical point

4x-100=0

x=25 m

Since y=50-x

y=25 m

And the total area is

[tex]A=25^2+25^2=1250\ m^2[/tex]

Note: if we set any other combination for x and y, say x=20 m and y=30m we would get greater areas

[tex]A=20^2+30^2=1300\ m^2[/tex]

The first option gives us the greatest area of 2500 m^2 and the second option has the least area of 1250 m^2

To get the greatest area with 200 meters of fencing, construct one single square enclosure, which will provide an area of 2500 m^2. For the least area, construct two separate square enclosures, which will give a total area of 312.5 m^2.

The question is asking how to use 200 meters of fencing to create square enclosures for the least and greatest area. For a given perimeter, a square has the largest area of any rectangle, so if we want to maximize the area enclosed, we would create one large square.

To find the size of the square, we would divide the total length of the fence by the number of sides of our square. So, 200 m / 4 sides = 50 m per side. Hence, the area of this square would be 50 m * 50 m = 2500 m^2.

For the least area, we would make 2 separate squares, each with half the fencing. So, each square would have 50 m of fence, or 12.5 m per side. The area for each square would then be 12.5 m * 12.5 m = 156.25 m^2, for a total area of 312.5 m^2 when you add both squares together.

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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).

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)