Use the Pythagorean theorem to determine which of the following give the measures of the legs and hypotenuse of a right triangle. Which apply. 3,4,5. Or. 4,11,14. Or. 9,14,17. Or 8,14,16. Or. 8,15,17

Answer:

67.84 km

Step-by-step explanation:

let her initial speed be 'u'

Time with speed u = 2 hours 50 minutes = 2.83 hours

Distance = speed × Time

therefore,

Distance = u × 2.83  ............(1)

when speed increase by 1 km/h

speed, v = u + 1

Time taken = 2 hours 50 minutes - 7 minutes

=  2 hours 43 minutes

= 2.7167 hours

Therefor,

Distance = ( u + 1 ) ×  2.7167 .............(2)

since, the distance in both the cases will be same

therefore,

from 1 and 2

u × 2.83 = ( u + 1 ) ×  2.7167

or

u × 1.0417 = u + 1

or

0.0417u = 1

or

u = 23.97 km/hr

Therefore,

Distance of loop = u × 2.83

= 23.97 × 2.83

= 67.84 km

The question involves calculating the distance of a loop that a bicyclist rides in two different conditions. Using speed, time and distance relation, we first set up two equations based on the given scenarios and solve them to find the distance. The distance turns out to be approximately 25.41 km.

This problem can be solved using the relationship between speed, distance and time. Let's denote the distance of the loop around Lake Washington as D km and her original average speed as S km/hr. Initially, she covers the distance in 2 hours and 50 minutes (which is equal to 2.83 hours when converted into hours).

So, we have the first equation as:  D = S x 2.83

Then, she increases her speed by 1 km/hr, and it reduces the total time by 7 minutes (which comes out to be 0.117 hours), resulting in a new time of 2.83 - 0.117 = 2.713 hours to cover the same distance. With her increased speed (S + 1 km/hr), this gives us the second equation as: D = (S + 1) x 2.713

Setting the equations equal to each other, we solve for S and subsequently find D. After solving, we get D approximately equal to 25.41 km.

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Answer:

[tex]\oint_C F \,dr = 0[/tex]

Step-by-step explanation:

In this problem, we have a vector field

                 [tex]F = \left \langle 2x-z^2,x^2+yz,-2xz +\frac{y^2}{2} \right \rangle[/tex].

We need to find the line integral

                                      [tex]\oint_C F \,dr[/tex]

where [tex]C[/tex] is a circle

                    [tex]r(t) = \left \langle 2 \cos t, 4\sin t, 5 \cos t \right \rangle, \quad 0 \leq t \leq 2 \pi[/tex].

As we can see, the vector filed [tex]F[/tex] is defined [tex]\forall (x,y,z) \in \mathbb{R}^3[/tex] and its component functions have continuous partial derivatives.

First, we need to find the curl of the vector filed [tex]F[/tex].

[tex]\text{curl} F = \begin{vmatrix}i & j & k\\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\2xy-z^2 & x^2 + yz & \frac{y^2}{2} - 2xz\\\end{vmatrix}[/tex]

Therefore,

[tex]\text{curl}F = i \left( \frac{\partial}{\partial y} \left( \frac{y^2}{2} - 2xz\right) - \frac{\partial}{\partial z} \left( x^2+yz\right)\right) - j \left( \frac{\partial}{\partial x} \left( \frac{y^2}{2} - 2xz\right) - \frac{\partial}{\partial z} \left( 2xy-z^2\right)\right)\\ \phantom{12356} +k \left( \frac{\partial}{\partial x} \left( x^2+yz\right) - \frac{\partial}{\partial y} \left( 2xy-z^2\right)\right)[/tex]

Now, we can easily calculate the needed partial derivatives and we obtain

[tex]\text{curl} F = i(y-y) - j(-2z + 2z) + k(2x-2x) = 0i + 0j+0k = \langle 0,0,0 \rangle[/tex]

So, the vector field [tex]F[/tex] is defined [tex]\forall (x,y,z) \in \mathbb{R}^3[/tex] , its component functions have continuous partial derivatives and [tex]\text{curl} F = 0[/tex] .Therefore, by a well-known theorem, [tex]F[/tex] is a conservative field.

Since [tex]C[/tex] is a closed path, we obtain that

                                           [tex]\oint_C F \,dr = 0[/tex]

Answer:

We can assume that the sample size is large enough to use the z distribution as an approximation of the t distribution

Now we need to find on the z distribution a value that accumulates 0.02 of the area on the left and this value is [tex]z_{crit}=-2.05[/tex]

We can use the following excel code to verify it: "=NORM.INV(0.02,0,1)"Reject H0 if test statistic < -2.05

And for our case our calculated value <-2.05 so we have enough evidence to reject the null hypothesis at 2% of significance.

Step-by-step explanation:

Data given and notation  

[tex]\bar X=13.8[/tex] represent the sample mea   n

[tex]s=0.3[/tex] represent the sample standard deviation

[tex]n=45[/tex] sample size  

[tex]\mu_o =68[/tex] represent the value that we want to test

[tex]\alpha=0.02[/tex] represent the significance level for the hypothesis test.  

t 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.  

We need to conduct a hypothesis in order to check if the mean is at least 14 oz, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 14[/tex]  

Alternative hypothesis:[tex]\mu < 14[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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{13.8-14}{\frac{0.3}{\sqrt{45}}}=-4.47[/tex]    

Critical region

We can assume that the sample size is large enough to use the z distribution as an approximation of the t distribution

Now we need to find on the z distribution a value that accumulates 0.02 of the area on the left and this value is [tex]z_{crit}=-2.05[/tex]

We can use the following excel code to verify it: "=NORM.INV(0.02,0,1)"

So then the correct answer for this case would be:

Reject H0 if test statistic < -2.05

And for our case our calculated value <-2.05 so we have enough evidence to reject the null hypothesis at 2% of significance.

The decision criterion to reject the null hypothesis that the mean weight is at least 14 oz at the 0.02 level of significance is to reject H0 if the test statistic is less than -2.33, as derived from the standard normal z-table.

To determine the criterion for rejecting the null hypothesis in a hypothesis test like the one presented - where a cereal company claims that the mean weight of the cereal in its packets is at least 14 oz - we use the level of significance (alpha). Since the level of significance in this problem is 0.02, we refer to the standard normal z-table to find the critical z-value that corresponds to this significance level for a one-tailed test, which in this case is a left-tailed test because we are checking if the mean weight is less than the claim (14 oz).

Using the z-table, we find that the critical z-value for an alpha of 0.02 in a one-tailed test is approximately -2.33. If our calculated test statistic is less than this critical value, we would reject the null hypothesis in favor of the alternative. Therefore, the decision criterion is: Reject H0 if test statistic < -2.33.

Since the given options are 'Reject H0 if test statistic < 2.05', 'Reject H0 if test statistic > -2.05', 'Reject H0 if test statistic < -2.05', and 'Reject H0 if test statistic < -2.33', the correct criterion for this test, considering the level of significance is 0.02, is 'Reject H0 if test statistic < -2.33'.

Answer:

4 pounds of of Vigoro Ultra Turf fertilizer and 1 pound of Parkera's Premium Starter will yield the required mixture.

Step-by-step explanation:

In bag of Vigoro Ultra Turf fertilizer:

29 pounds of nitrogen, 3 pounds of phosphoric acid, and 4 pounds of potash

In bag of Parkera's Premium Starter:

18 pounds of nitrogen, 25 pounds of phosphoric acid, and 6 pounds of potash

Let the mass of Vigoro Ultra Turf fertilizer = x

Amount of nitrogen in x amount = 29 x

Amount of phosphoric acid  in  x amount = 3x

Let the mass of Parkera's Premium Starter= y

Amount of nitrogen in y amount = 18 y

Amount of phosphoric acid  in y  amount = 25 y

Amount of nitrogen in desired mixture formed by combing x and y fertilizers bag : 134 pounds

29 x + 18 y = 134 ..[1]

Amount of phosphoric acid  in desired mixture formed by combing x and y fertilizers bag : 37 pounds

3x + 25 y = 37 ..[2]

Solving [1] anf [2] by substitution method :

[tex]x=\frac{134-18y}{29}[/tex]

[tex]3\times \frac{134-18y}{29}+25y=37[/tex]

y = 1

[tex]x=\frac{134-18y}{29}=4[/tex]

4 pounds of of Vigoro Ultra Turf fertilizer and 1 pound of Parkera's Premium Starter will yield the required mixture.

To create a mixture containing 134 pounds of nitrogen, 37 pounds of phosphoric acid, and 22 pounds of potash, 400 pounds of Vigoro Ultra Turf and 100 pounds of Parker's Premium Starter are required, based on solving a system of linear equations.

We are tasked with finding the amount of two different fertilizers needed to obtain a mixture with 134 pounds of nitrogen, 37 pounds of phosphoric acid, and 22 pounds of potash. The two fertilizers are Vigoro Ultra Turf and Parkerâ's Premium Starter. We'll use a system of linear equations to solve this problem.

Let V represent the amount (in pounds) of Vigoro Ultra Turf and P represent the amount (in pounds) of Parkerâ's Premium Starter. The equations based on the given information are:

Now we solve these equations simultaneously to determine V and P. After solving, we find that V = 400 and P = 100. Thus, we need 400 pounds of Vigoro Ultra Turf and 100 pounds of Parkerâ's Premium Starter to obtain the desired mixture.

Answer:

a) No Unique solution

b) No Unique solution

c) No unique solution

Step-by-step explanation:

The step by step explanation is as shown in the attachment below

Answer:

29C3 = 3654

Step-by-step explanation:

What is applied here is combination that n - combination of r which is a subscript and the n - is a superscript.

Attached is the detailed explanation.

Answer:

[tex]\frac{1}{7}[/tex]

Step-by-step explanation:

The table containing information about the sexes of twins and and the type of twins are given in the table below with corresponding weightage.

Now, since we know that she is expecting twins, the probability of the event

                [tex]A - '\text{the woman will have identical twins }'[/tex]

is calculated as follows

[tex]Pr(A) = \dfrac{\text{total number of boy/boy, boy/girl, girl/boy and girl/girl identical twins}}{\text{total number of both identical and frathernal twins}}[/tex]

By using the informations from the given table, we obtain

                                        [tex]Pr(A) = \frac{2 + 0 + 0 +2}{28} = \frac{4}{28} = \frac{1}{7}[/tex]

Answer: Yes, you are correct it's $8.95 because like they stated in the word problem p= the price of each bowl and like they said in the word problem they charge $8.95 a bowl. So, $8.95 is the answer.

Step-by-step explanation:

Answer:

Red Marble is the color in the bag

Step-by-step explanation:

Total number of marbles ; 201

number of red marbles ; 101

number of blue marbles ; 101

first possibility ; From what is said as the condition, it can be inferred that the number of blue marbles is reduced by a factor of 1 , while the number of red marbles remains constant. this shows that the two marbles of the same color that was removed is most likely BLUE.

Second possibility ; Two red marbles are added (of the same color) and add a blue marbles, implies the the red marbles reduce by a factor of 2 while the blue marbles increase by a factor of 1. this is also possible.

Third possibility ; Two marbles of different color implies 1R and 1B , and then a red marble is added. this shows that the number of red marble still remains constant ( 1R removed, 1R added), while the number of blue reduce by 1 ( remove 1B, and not replaced back).

From the various possibilities, it can be inferred that the number of red marbles is either increasing by a factor of 2 or remains constant. this further shows that in all three , 1R marble will always remain in the bag no matter the possibility as such in all three scenerios, it further shows that if there is only a marble remaining, it would have been the RED MARBLE WHICH IS AUTOMATICALLY THE LAST MARBLE from the analysis done above.

Answer:

P = 10/171

P = 0.058

The probability that both fish selected are Catfish is 0.0058

Step-by-step explanation:

Given:

The tank contains;

Algae eaters = 8

Catfish = 5

Tilapia fish = 6

Total number of fish = 8+5+6= 19

To determine the probability that both fish selected are Catfish P;

P = 5/19 × 4/18

( The number of the total number of fishes, and Catfish reduce by one after each selection because it is selected without replacement)

P = 10/171

P = 0.058

Answer:

Step-by-step explanation:

Given that a die is thrown twice.  

X1, X2 are the outcomes in 2 throws

X1 = minimum of two

X1 can be either 1 or 2...6

Total outcomes are 36.

For x1 =1, fav ourable outcomes are (1,1) (1,2)...(1,6) (6,1)...(2/,1)= 11

P(X1=1) = [tex]\frac{11}{36}[/tex]\

P(X1=2) =Prob for one die showing two and other die showing 2 to 6

= [tex]\frac{9}{36}[/tex]

P(x1=3) = Prob for one die showing three and other die showing 3 to 6

=[tex]\frac{7}{36}[/tex]

thus we find that probability is reducing by 2 in the numerator

P(x1=4) = 5/36 followed by 3/36 for 5 and 1/36 for 6

Final answer:

The random variable X represents the minimum outcome when rolling a die twice. The values of X range from 1 to 6 and its distribution is given.

Explanation:

The random variable X represents the minimum outcome when rolling a die twice. In other words, X is the smaller of the two outcomes.

X can take on values from 1 to 6 since the outcomes of rolling a die are integers between 1 and 6. The minimum value of X is 1, which occurs when both dice show a 1.

The distribution of X is as follows:

Answer:

a) (-∞,-9)∪(-9,9)∪(9,+∞)

b) (-∞,+∞)

Step-by-step explanation:

The domain of a real function is the largest subset og the real line in which it is defined.

a) The function [tex]f(x)=\frac{81-e^{x^2}}{1-e^{81-x^2}}[/tex] is defined for all values of x in which the denominator is not zero. The denominator is zero if [tex]0=1-e^{81-x^2}[/tex], that is,  [tex]e^{81-x^2}=1[/tex]. The exponential function is equal to one only if the exponent is equal to zero, then the values of x that nullify the denominator satisfy [tex]81-x^2=0[/tex], thus x=9 or x=-9. Then the domain of f is the set of all real numbers such that x≠9 and x≠-9, which is the set (-∞,-9)∪(-9,9)∪(9,+∞).

b) In this case the denominator is [tex]e^{\cos x}[/tex] which is always positive. Thus the denominator is not zero for all real x, then the fomain is the real line, which is the interval (-∞,+∞).

Final answer:

The domain of the function f(x) = (81-e^x^2)/(1-e^81-x^2) is (-∞, -9) ∪ (-9, 9) ∪ (9, ∞). The domain of the function f(x) = 7 + x/e^cos(x) is (-∞, ∞).

Explanation:

(a) To find the domain of the function f(x) = (81-e^x^2)/(1-e^81-x^2), we need to determine the values of x for which the function is defined. The function is defined as long as the denominator is not zero. So, we need to solve the equation 1 - e^(81-x^2) = 0 to find where the denominator equals zero. Simplifying this equation, we get e^(81-x^2) = 1. Taking the natural logarithm of both sides, we have 81-x^2 = 0. Therefore, x^2 = 81, and x = ±9. The domain of the function is (-∞, -9) ∪ (-9, 9) ∪ (9, ∞).

(b) To find the domain of the function f(x) = 7 + x/e^cos(x), we need to determine the values of x for which the function is defined. The function is defined as long as the denominator e^cos(x) is not zero. Since the exponential function is always positive, we know that e^cos(x) will never be zero. Therefore, the domain of the function is the set of all real numbers, (-∞, ∞).

Answer:

First Part:

exponential function [tex]e^t[/tex] is the one whose first order derivative is the function itself.

Second Part:

[tex]y=ce^{At}\\y'=Ay[/tex]

Third Part:

[tex]y=ce^t\\y'=ce^t\\y'=y[/tex]

Step-by-step explanation:

First Part:

In calculus exponential function [tex]e^t[/tex] is the one whose first order derivative is the function itself.

Where:

t is independent variable.

Derivative is represented as:

[tex]y=ce^t\\y'=\frac{d(ce^t)}{dt} \\y'=ce^t\\y'=y[/tex]

Where:

c is any number.

Second Part:

Consider the constant A.

The function will become:

[tex]y=ce^{At}\\y'=\frac{d(ce^{At})}{dt} \\y'=cAe^{At}\\y'=Ay[/tex]

Third Part:

Derivative is represented as:

[tex]y=ce^t\\y'=\frac{d(ce^t)}{dt} \\y'=ce^t\\y'=y[/tex]

Where:

c is any number.

Final answer:

The differential equation is [tex]\( \frac{df}{dx} = Cf(x) \),[/tex] with the solution[tex]\( f(x) = Ce^x \), where \( C \)[/tex] is a constant.

Explanation:

The function that satisfies both conditions is [tex]\( f(x) = Ce^x \), where \( C \)[/tex] a constant. Its first derivative is[tex]\( f'(x) = Ce^x \),[/tex]which is a constant multiple of itself. This can be expressed as a first-order differential equation:

[tex]\[ \frac{df}{dx} = Cf(x) \][/tex]

with the solution:

[tex]\( f(x) = Ce^x \), where \( C \)[/tex]

This differential equation represents exponential growth or decay, depending on the sign of [tex]\( C \). For \( C > 0 \)[/tex], the function exponentially grows as [tex]\( x \) increases, while for \( C < 0 \)[/tex]it exponentially decays. The solution captures phenomena like population growth, radioactive decay, or compound interest, making it a fundamental concept in calculus and applied mathematics.

Answer:

For First Solution: [tex]y_1(t)=e^t[/tex]

[tex]y_1(t)=e^t[/tex] is the solution of equation y''-y=0.

For 2nd Solution:[tex]y_2(t)=cosht[/tex]

[tex]y_2(t)=cosht[/tex]  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: [tex]y_1(t)=e^t[/tex]

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

[tex]y_1(t)=e^t[/tex]

First order derivative:

[tex]y'_1(t)=e^t[/tex]

2nd order Derivative:

[tex]y''_1(t)=e^t[/tex]

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence [tex]y_1(t)=e^t[/tex] is the solution of equation y''-y=0.

For 2nd Solution:

[tex]y_2(t)=cosht[/tex]

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

[tex]y_2(t)=cosht[/tex]

First order derivative:

[tex]y'_2(t)=sinht[/tex]

2nd order Derivative:

[tex]y''_2(t)=cosht[/tex]

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence [tex]y_2(t)=cosht[/tex]  is the solution of equation y''-y=0.

The functions [tex]y(t)=e^t[/tex] and [tex]y(t)=cosht[/tex] are solutions of given differential equation.

We have to prove that given function is a solution of the differential equation.

Given function,   [tex]y(t)=e^t[/tex]

Given differential equation are, [tex]y''-y=0[/tex]

  So that,  [tex]y'(t)=e^t , y''(t)=e^t[/tex]

Substitute values in given differential equation.

                    [tex]e^t - e^t=0[/tex]

Thus, function [tex]y(t)=e^t[/tex] is solution of given differential equation.

Another function is,  [tex]y(t)=cosht[/tex]

So that,  [tex]y'(t)=sinht , y''(t)=cosht[/tex]

Substitute values in given differential equation.

                    [tex]cosht-cosht=0[/tex]

Thus, function [tex]y(t)=cosht[/tex] is solution of given differential equation.

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Answer:

Qualitative variable

Step-by-step explanation:

The response variable is qualitative because it can be categorize as "Good time" or "Bad time". The time is also rated as "Good" or "Bad" for taking a lengthy vacation and so it involves ordinal scale of measurement.

We can say that our response variable is qualitative and ordinal. In short response variable  is of qualitative type of variable and the scale of measurement it involves is ordinal scale.

Answer: Sample size n is 9.

The sample mean M is 84.

Step-by-step explanation:

Given : A professor gives a statistics exam.

The exam has 100 possible points.

The scores for the students in the second classroom are as follows:

88 84 84 52 100 84 92 84 88

Since , the number of scores of different students= 9

∴ Sample size is n= 9.

Now , Sample mean = [tex]\dfrac{\text{Sum of all data values}}{\text{Sample size}}[/tex]

[tex]=\dfrac{88+ 84 +84 +52 +100+ 84+ 92+ 84+ 88 }{9}\\\\=\dfrac{756}{9} =84[/tex]

Hence, the sample mean M is 84.

We have that the sample size, n, and the sample mean, M. is mathematically given as

From the question we are told

A professor gives a statistics exam.

Generally the equation for the  Mean is mathematically given as

[tex]M=\sum/n\\\\\Therefore\\\\M=\frac{88 +84 +84+ 52+ 100+ 84+ 92+ 84+ 88 }{9}\\\\M=84[/tex]

Therefore

New mean

M_n=84+4

M_n=88

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Answer:

Roughly the same as what?

Step-by-step explanation:

Answer:

A. 400 GRAMS

B. 120 GRAMS

C. 356 GRAMS

D. 340 GRAMS

PENN FOSTER

ANSWER IS D 340 GRAMS

Step-by-step explanation:1 ounce = 28.3495 12 ounces = 28.3495 * 12 = 340.194 grams

The time needed to run the second half of the race is 8.44 s

Step-by-step explanation:

The total distance covered by the athlete during the race is

d = 200 m

And the total time taken to cover this distance is

T = 19.19 s

We also know that the time the athlete needs to cover the first half of the race is

[tex]t_1 = 10.75 s[/tex]

Also, we know that

[tex]T=t_1 + t_2[/tex]

where [tex]t_2[/tex] is the time the athlete takes to cover the second half of the race.

Re-arranging this equation and susbtituting the values, we find the value of t2:

[tex]t_2 = T-t_1 = 19.19-10.75=8.44 s[/tex]

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Answer:

Option B is right

Step-by-step explanation:

P(A or B)

In set theory if A and B are two sets we have either A occurs or B occurs or both as

P(A or B) = P(AUB)

This implies that A occurs or B occurs

Probability is calculated as P(A)+P(B)-P(AB)

P(AB) here represents both occurring and this is subtracted as this was added two times

Option A is wrong because this is AB

Option C is wrong because it is not necessary A occurs first

D is wrong because this is A'UB' where A' represents the complement of A

So correct answer out of four options is

B. the probability that in a single​ trial, event A​ occurs, event B​ occurs, or they both occur.

The notation P(A or B) in probability specifies the probability that in a single trial, either event A happens, event B happens, or they both happen.

The notation P(A or B) in probability theory denotes the probability that event A occurs, event B occurs, or both events occur in a single trial. If we look at the four options provided, option B fits this definition. Hence, P(A or B) indicates the probability that in a single trial, event A occurs, event B occurs, or they both occur. It does not imply that both events A and B have to occur concurrently in the same trial (option A), nor that event A has to occur in one trial followed by event B in another trial (option C), and certainly not the probability that event A or event B does not occur in a single trial (option D).

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Answer:

a) 5 units/s

b) yes

c) counter-clockwise

d) yes

Step-by-step explanation:

part a

r(t) = cos (5t) i + sin (5t)j

v(t) = dr(t) / dt = -5sin(5t) i + 5cos(5t)j

[tex]mag( v(t)) = \sqrt{(-5sin(5t))^2 + (5cos(5t))^2} \\mag( v(t)) = \sqrt{25sin^2(5t) + 25cos^2(5t)} \\ \\mag( v(t)) = \sqrt{25*(sin(5t)^2 + cos(5t)^2)} \\\\mag( v(t)) = \sqrt{25} \\\\mag( v(t)) = 5 units/s[/tex]

Hence, the particle has a constant speed of 5 units/s

part b

a(t) = dv(t) / dt = -25cos(5t) i - 25sin(5t)j

To check orthogonality of two vectors their dot product must be zero

a(t) . v(t) = (-25cos(5t) i - 25sin(5t)j) . (-5sin(5t) i + 5cos(5t)j)

= 125cos(5t)*sin(5t) -125cos(5t)*sin(5t)

= 0

Yes, the particles velocity vector is always orthogonal to acceleration vector.

part c

Use any two values of t and compute results of r(t)

t = 0 , r(0) = 1 i

t = pi/2, r(0) = j

Hence we can see that the particle moves counter-clockwise

part d

Find the value r(t) at t=0

r(0) = cos (0) i + sin (0) j

r(0) = 1 i + 0 j

Yes, the particle starts at point ( 1, 0)

Answer:

The probability that they get a sampling error of greater than 1.0 mg/L is 0.04.

To calculate this value, I assume that the samples are randomly collected

Step-by-step explanation:

Sampling error can be calculated using the formula

[tex]\frac{t*s}{\sqrt{n}}[/tex]  where

For the sampling error of 1.0mg/L we have

[tex]1=\frac{t*1.60}{\sqrt{10}}[/tex]    

Solving for t we have t≈1.976

Then the probability that they get a sampling error of greater than 1.0 mg/L is

P(t>1.976) ≈ 0.04.

In other words, we are 96% confident that the sampling error is within 1.0 mg/L.

To calculate this value, I assume that the samples are randomly collected.

Answer:

i) -28 ft/s

ii) -21.6 ft/s

iii) -20.8 ft/s

iv) -20.16 ft/s

b)  -20 ft/s

Step-by-step explanation:

if v represents vertical velocity then if the height y is

y = 44*t − 16*t²

and the instant velocity v is the derivative with respect to the time

v= dy/dt = 44*(1) - 16*(2*t) = 44- 32*t

while the average velocity is va= (y-y₀)/(t-t₀)

where t₀ = 2 and y₀=y(t₀) =  44*2 − 16*2² = 24

then

va= (y-y₀)/(t-t₀) = (44*t − 16*t² -  24)/(t-2)

for

i) t= 0.5s + 2 s= 2.5 s

va = (44* 2.5  − 16* 2.5 ² -  24)/( 2.5 -2) = -28 ft/s

ii) t= 0.1s + 2 s= 2.1 s

va = (44* 2.1  − 16* 2.1 ² -  24)/( 2.1 -2) = -21.6 ft/s

iii) t= 0.05 s + 2 s= 2.05 s

va = (44* 2.05  − 16* 2.05 ² -  24)/( 2.05 -2) = -20.8 ft/s

iv) t= 0.01s + 2 s= 2.01 s

va = (44* 2.01  − 16* 2.01 ² -  24)/( 2.01 -2) = -20.16 ft/s

b) the instantaneous velocity when t=2

v (t=2) =  44- 32*(2) = -20 ft/s

Answer:

i) The first  -28 ft/s

ii) Then second -21.6 ft/s

iii) Thirds one is -20.8 ft/s

iv) Then  -20.16 ft/s

b)  -20 ft/s

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Answer:

There is a 99.99998% probability that at least one valve opens.

Step-by-step explanation:

For each valve there are only two possible outcomes. Either it opens on demand, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

[tex]n = 5, p = 0.93[/tex]

Calculate P(at least one valve opens).

This is [tex]P(X \geq 1)[/tex]

Either no valves open, or at least one does. The sum of the probabilities of these events is decimal 1. So:

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{5,0}.(0.93)^{0}.(0.07)^{5} = 0.0000016807[/tex]

Finally

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000016807 = 0.9999983193[/tex]

There is a 99.99998% probability that at least one valve opens.

Answer:

0.930000

Step-by-step explanation:

Answer:

8 points below the mean.

Step-by-step explanation:

Let 'X' be the score of the third person, and let 'M' be the mean score of the sample.

If the other two people scored 5 and 3 above the mean, in order to maintain consistency, the following expression must be true:

[tex]M+5+(M+3)+X =3M\\X=3M-2M-8\\X=M-8[/tex]

Therefore, the third person has a score 8 points below the mean.

Given that the other two people's scores are collectively 8 points above the mean, the third person's score must be 8 points below the mean to balance out the total deviation.

This question is about determining the score of the third person in a sample relative to the mean of the sample. We know that the total deviation from the mean must equal zero because the mean is the average of all the scores. Given the first person has a score 5 points above the mean, and the second person has a score 3 points above the mean, the total deviation above the mean is (5+3)=8 points. For the score deviations to balance out to zero, the third person's score must be 8 points below the mean.

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Answer:

1) [tex] y =4x^2 +7[/tex]

2) [tex] y =7s^2 -3s^4 +181[/tex]

Step-by-step explanation:

Assuming that our function is [tex] y = f(x)[/tex] for the first case and [tex] y=f(s)[/tex] for the second case.

Part 1

We can rewrite the expression like this:

[tex] \frac{dy}{dx} =8x[/tex]

And we can reorder the terms like this:

[tex] dy = 8 x dx[/tex]

Now if we apply integral in both sides we got:

[tex] \int dy = 8 \int x dx[/tex]

And after do the integrals we got:

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

Now we can use the initial condition [tex] y(0) =7[/tex]

[tex] 7 = 4(0)^2 +c, c=7[/tex]

And the final solution would be:

[tex] y =4x^2 +7[/tex]

Part 2

We can rewrite the expression like this:

[tex] \frac{dy}{ds} =14s -12s^3[/tex]

And we can reorder the terms like this:

[tex] dy = 14s -12s^3 dx[/tex]

Now if we apply integral in both sides we got:

[tex] \int dy = \int 14s -12s^3 ds[/tex]

And after do the integrals we got:

[tex] y = 7s^2 -3s^4 +c[/tex]

Now we can use the initial condition [tex] y(3) =1[/tex]

[tex] 1 = 7(3)^2 -3(3)^4 +c, c=1-63+243=181[/tex]

And the final solution would be:

[tex] y =7s^2 -3s^4 +181[/tex]

Answer:

Distance between the van and Morristown is changing at the rate of 13.22 miles per hour.

Step-by-step explanation:

From the figure attached,

Van starts from C (City of Morristown), reaches the point A 191 miles due North and then it travels with a speed of 25 miles per hour due East from A towards B.

We have to calculate the rate of change of distance BC, when the van reaches point B which is 119 miles away from A.

By Pythagoras theorem in the triangle ABC,

[tex]BC^{2}=AB^{2}+AC^{2}[/tex]

Distance AC is constant equal to 191 mi.

By differentiating the equation with respect to time 't'

[tex]2BC.\frac{d(BC)}{dt}=0+2AB.\frac{d(AB)}{dt}[/tex]

[tex]BC.\frac{d(BC)}{dt}=AB.\frac{d(AB)}{dt}[/tex]

Since BC² = (119)²+ (191)²

BC = √50642 = 225.04 miles

From the differential equation,

[tex](225.03).\frac{d(BC)}{dt}=119\times 25[/tex] [Since [tex]\frac{d(AB)}{dt}=25[/tex] miles per hour]

[tex]\frac{d(BC)}{dt}=13.22[/tex] miles per hour

Therefore, distance between the van and Morristown is changing at the rate of 13.22 miles per hour.

Answer:

54,18

Step-by-step explanation:

Let one of the numbers be x. The other number be represented as 36-x

(x-(36-x ) = 36

open brackets

x - 36 + x  =  36

x + x = 36+ 36 = 72

2x = 72

x = 72/2 = 36

x =  36

The product can then be represented as y = x(36-x) or y=36x-x²

 The maximum or minimum is always on the axis of symmetry which has the formula x=-b/2a.

In our case, the axis of symmetry is -36/-2, so x=18.

If one number is 18 and the 2 numbers differentiate by 36, the other number is 18 + 36 = 54

So the 2 numbers are 18 and 54 and the minimum product is 972

The two numbers are 18 and 54 and the minimum product is 972

If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.

Let suppose one of the numbers be x and the other number be represented as 36-x

Then the equation would be;

(x-(36-x ) = 36

Now open brackets;

x - 36 + x  =  36

x + x = 36+ 36 = 72

2x = 72

x = 72/2 = 36

x =  36

Therefore, The product can be represented as y = x(36-x) or y=36x-x²

Here maximum or minimum is always on the axis of symmetry which is the formula x=-b/2a.

In our case; the axis of symmetry is -36/-2,

Then x=18.

If one number is 18 and the 2 numbers differentiate by 36 then the other number must be 18 + 36 = 54

Hence the two numbers are 18 and 54 and the minimum product is 972

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Answer:

635599482kg

Step-by-step explanation:

A gourmet chocolate candy has 7.00 gg of dietary fat in each 22.7-gg piece

for the 22.7 piece , the gg is 7 x 22.7  = 158.9gg in total

1gg = 2204622.6218488 lbs

158.9gg in total  = 158.9 x 2204622.6218488 = 350314534.612

for the box of 4.00 lblb of candy

it is 350314534.612 x 4 = 1401258138.45lb

since our answer is needed in kilograms we convert 1401258138.45lb to kilograms

1lb = 0.453592kg

1401258138.45lb =  1401258138.45lb x 0.453592kg = 635599481.513 ≈ 635599482kg

Answer:

[tex] (\forall x) S(x)\rightarrow A(x) [/tex]

[tex] (\exists x)\neg S(x)\wedge A(x) [/tex]

Step-by-step explanation:

Everyone who studied for the test received an A on the test.

That means- if you studied for the test you will recived an A, and it is hold for everyone, so we will use quantifire [tex]\forall[/tex].

[tex] (\forall x) S(x)\rightarrow A(x) [/tex]

It means: for every student holds- If it is correct that student x studied then  the student got an A.

Someone who did not study for the test received an A on the test.

It means that, there is at least one student hwo didn't studie but student got an A. So we have conjuction of two sentences (negation of the S(x) and A(x) for some student- for that we use existential quantifie).

[tex] (\exists x)\neg S(x)\wedge A(x) [/tex]

Final answer:

The logical expression involving predicates S(x) and A(x) can be represented by ∀x(S(x) → A(x)), stating that all students who studied for the test received an A.

Explanation:

The question involving the predicates S(x): x studied for the test and A(x): x received an A on the test revolves around predicate logic, where we aim to understand and analyze the logical relations of sentences with subjects and predicates within a specified domain of discourse.

To define the logical expression that describes a relationship between studying for a test and receiving an A would depend on the specific relationship we want to express. For instance, a possible logical expression could be ∀x(S(x) → A(x)), which translates to 'For all students x in the class, if x studied for the test, then x received an A on the test.'

Final answer:

To evaluate the manufacturer's claim, a one-sample t-test can be conducted. The test statistic can be calculated using the given information.

Explanation:

To conduct a hypothesis test to evaluate the manufacturer's claim, we can use a one-sample t-test. The null hypothesis (H0) is that the mean MPG is equal to 39.0, and the alternative hypothesis (Ha) is that the mean MPG is not equal to 39.0. We can calculate the test statistic using the formula:

test statistic = (sample mean - population mean) / (standard deviation / sqrt(sample size))

Using the given values:

sample mean = 38.8

population mean = 39.0

standard deviation = 2.2

sample size = 120

Substituting these values, the test statistic is:

test statistic = (38.8 - 39.0) / (2.2 / sqrt(120))