Do not round intermediate calculations. round your answers to two decimal places. a. define a random variable number of times that owner-occupied units had a water supply stoppage lasting or more hours in the past months and develop a probability distribution for the random variable. (let represent or more times.) 0.04 0.34 0.41 0.17 0.04 1 b. compute the expected value and variance for . 1.82 c. define a random variable number of times that renter-occupied units had a water supply stoppage lasting or more hours in the past months and develop a probability distribution for the random variable. (let represent or more times.) 0 0.03 0.22 0.52 0.23 1 d. compute the expected value and variance for . e. what observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units

Answer:

6k-5

Step-by-step explanation:

8k-(5+2k) write the equation out

8k-5-2k distribute the (-) to everything inside the parenthesis

6k-5 combine like terms

The answer is option d. -1.865, as it is the value that satisfies P(Z ≤ b) = 0.0311. The other options (-1.87, -1.86, -1.8) do not correspond to the given cumulative probability.

In this scenario, P(Z ≤ b) represents the cumulative probability of a standard normal distribution up to the value of b. To find the corresponding value of b, we need to find the z-score that corresponds to a cumulative probability of 0.0311.

By looking up the z-table or using a statistical calculator, we can find that the z-score corresponding to a cumulative probability of 0.0311 is approximately -1.865.

Therefore, the answer is option d. -1.865, as it is the value that satisfies P(Z ≤ b) = 0.0311. The other options (-1.87, -1.86, -1.8) do not correspond to the given cumulative probability.

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Therefore, in a 2x2 table, the pattern of cell frequencies that would indicate independence is d. All cell frequencies are exactly the same.

In a 2x2 contingency table, the expected cell frequencies under the null hypothesis of independence are equal for all cells. If the observed cell frequencies in the table are approximately equal to the expected cell frequencies, then we can conclude that there is no significant association between the two variables being studied. In other words, the pattern of observed cell frequencies is consistent with the null hypothesis of independence. Therefore, if all cell frequencies are exactly the same, it suggests that the variables are independent, as each cell has an equal chance of being filled by any observation regardless of the value of the other variable.

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Answer:

A = 5 + 3t

B = 7 + 2t

2 hours

Step-by-step explanation:

7 + 2t > 5 +3t

                      (subtract 2t from both sides)

7         > 5 + t

                      (subtract 5 from both sides)

2         > t

therefore in 2 hours, the parking would be cheaper in Garage A than B

Answer:

1/3 x 12=4

Step-by-step explanation:

The correct answer for 1/3*12 is 4.

1: Times 1/3 by 12

Answer: 4

Hope this helps.

The error which Loren made was C. she should have solved 149 = startfraction 19 over 9 endfraction (10) b for b.

This refers to those miscalculations or misjudgments about a situation or a mathematical solution to give wrong values.

Hence, because Loren is solving an equation of 0 = startfraction 19 over 9 endfraction (149) b for b to find the equation of a trend line that passes through the points (1, 130) and (10, 149), the error she made was option C.

This means that using point (10,149) which takes the form (x,y) the equation will be: 149=19/9(10)+b

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Answer:

C

Step-by-step explanation:

The correct set of possible results would be (0, 1, 0, 0), (1, 2, 1, 0) and (1, 2, 0, 1).

Your approach seems to be correct, but there seems to be a minor mistake in your final list of possible solutions. Let's go through the steps to clarify.

Given the initial conditions z=t=0, you obtained a partial solution (0,1,0,0).

Next, you solved the homogeneous equations for z=1 and t=0, which resulted in a solution (1,1,1,0).

Similarly, solving the homogeneous equations for z=0 and t=1 gives another solution (1,1,0,1).

To find the general solution, you combine the partial solution with the solutions obtained in the previous step, using parameters a and b.

(0,1,0,0) + a(1,1,1,0) + b(1,1,0,1)

Expanding this expression, you get:

(0+a+b, 1+a+b, 0+a, 0+b)

Simplifying, you obtain the following set of solutions:

(0, 1, 0, 0)

(1, 2, 1, 0)

(1, 2, 0, 1)

Therefore, the correct set of possible results would be:

(0, 1, 0, 0)

(1, 2, 1, 0)

(1, 2, 0, 1)

Note that (0, 1, 1, 1) is not a valid solution in this case, as it does not satisfy the initial condition z = 0.

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Answer:

smaller ∡ = 26°

larger ∡ = 154°

Step-by-step explanation:

opposite angles are congruent

adjacent angles are supplementary

in this problem, the angles are supplementary

'x' = measure of smaller angle

x + 50 + 4x = 180

5x = 130

x = 26

The given information and the relationship between the angles of a

parallelogram can be used to find the interior angles.

Reasons:

A property of a parallelogram is that the diagonally opposite angles are

equal, and the adjacent angles are supplementary.

Therefore, two of the four interior angles of a parallelogram are equal.

Let x and y represent the two angles of different measure in the

parallelogram

y = 4·x + 50°

y + x = 180°

Which gives;

y = 180° - x

Therefore;

180° - x = 4·x + 50°

180° - 50° = 4·x + x

130° = 5·x

x = 130° ÷ 5 = 26°

y = 4 × 26° + 50° = 154°

Therefore;

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a. Analytically, the integral evaluates to

\(I = 2x - (1/2)x^2 - (1/5)x^5 + (1/3)x^3 + (1/6)x^6 + C.\)

b. Using the trapezoidal rule, I = 0.3.

c. Using the composite trapezoidal rule with n = 2, I = 0.425. With n = 4, I = 0.353125.

d. Using Simpson's 1/3 rule, I = 0.33125.

e. Using Simpson's 3/8 rule, I = 0.34825.

f. The true percent relative error can be calculated based on the result from part a.

g. MATLAB calculations can be used to support the results and compare the different numerical methods.

a. To evaluate the integral analytically, we integrate term by term, and add the constant of integration, denoted as C.

b. The trapezoidal rule approximates the integral using trapezoids. For a single application, we evaluate the function at the endpoints of the interval and use the formula I = (b-a) * (f(a) + f(b)) / 2.

c. The composite trapezoidal rule divides the interval into smaller subintervals and applies the trapezoidal rule to each subinterval.

With n = 2, we have two subintervals, and with n = 4, we have four subintervals.

d. Simpson's 1/3 rule approximates the integral using quadratic interpolations. We evaluate the function at three equally spaced points within the interval and use the formula

I = (b-a) * (f(a) + 4f((a+b)/2) + f(b)) / 6.

e. Simpson's 3/8 rule approximates the integral using cubic interpolations. We evaluate the function at four equally spaced points within the interval and use the formula

I = (b-a) * (f(a) + 3f((2a+b)/3) + 3f((a+2b)/3) + f(b)) / 8.

f. The true percent relative error can be calculated by comparing the result obtained analytically with the result obtained numerically, using the formula: (|I_analytical - I_numerical| / |I_analytical|) * 100%.

g. MATLAB calculations can be performed to evaluate the integral using the different numerical methods and compare the results. The calculations will involve numerical approximations based on the given function and the specified methods.

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it will take Bob approximately 47.3 years to save $200,000 for his retirement.

How to find?

To determine the time it will take Bob to reach his retirement goal of $200,000, we can use the formula for compound interest:

A = P(1 + r/n)²(nt)

where:

A = the final amount

P = the initial principal (deposit)

r = the annual interest rate (as a decimal)

n = the number of times the interest is compounded per year

t = the time (in years)

In this case, we have:

P = $400 (monthly deposit)

r = 1.8% (annual interest rate, compounded monthly)

n = 12 (compounded monthly)

A = $200,000

We can solve for t by substituting the given values and solving for t:

200,000 = 400(1 + 0.018/12)²(12t)

500 = (1 + 0.0015)²(12t)

log(500) = 12t log(1.0015)

t = log(500) / (12 log(1.0015))

t ≈ 47.3

Therefore, it will take Bob approximately 47.3 years to save $200,000 for his retirement.

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Answer:height

Step-by-step explanation:

Answer:

area

Step-by-step explanation:

its right, trust

The P-value for the hypothesis test is approx. 0.7064.

To find the P-value for the hypothesis test, we need to perform a two-sample proportion test. The formula to calculate the test statistic for this test is given by:

\(\[ z = \frac{{(\hat{p}_1 - \hat{p}_2) - 0}}{{\sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{{n_1}} + \frac{1}{{n_2}}\right)}}} \]\)

where \(\hat{p}_1\) and \(\hat{p}_2\) are the sample proportions, \(\hat{p}\) is the combined sample proportion, \(n_1\) and \(n_2\) are the sample sizes, and 0 is the hypothesized difference in proportions.

In this case, we have \(n_1 = 50\), \(x_1 = 8\) (number of successes in sample 1), \(n_2 = 50\), and \(x_2 = 7\) (number of successes in sample 2).

First, we calculate the sample proportions:

\(\[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{8}{50} = 0.16 \]\)

\(\[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{7}{50} = 0.14 \]\)

Next, we calculate the combined sample proportion:

\(\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{8 + 7}{50 + 50} = 0.15 \]\)

Then, we calculate the standard error:

\(\[ SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{{n_1}} + \frac{1}{{n_2}}\right)} = \sqrt{0.15(1 - 0.15)\left(\frac{1}{50} + \frac{1}{50}\right)} \approx 0.0529 \]\)

Finally, we calculate the test statistic:

\(\[ z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{SE} = \frac{(0.16 - 0.14) - 0}{0.0529} \approx 0.3772 \]\)

To find the P-value, we look up the test statistic in the standard normal distribution table. In this case, the P-value is the probability of observing a test statistic greater than 0.3772 or less than -0.3772.

Consulting the table, we find that the P-value is approximately 0.7064. Therefore, the P-value for the hypothesis test is approximately 0.7064.

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The value of expression (a + b + c + d) is 20

Consider given equations x + y = 4       ........(1)

and 3xy = 4        .........(2)

The solutions of given equations are of the form \($x = \frac{a \pm b\sqrt{c}}{d}$\)

where a, b, c, and d are positive integers and the expression is completely simplified.

From equation (2),

y = 4/3x

Substitute above value of y in equation (1),

x + y = 4

x + (4/3x) = 4

3x² + 4 = 12x

3x² - 12x + 4 = 0

After solving above quadratic equation we get,

\(x=\frac{6\pm 2\sqrt{6} }{6}\)

Comapring this solution with \($x = \frac{a \pm b\sqrt{c}}{d}$\) we get,

a = 6, b = 2, c = 6 and d = 6

Now we find the value of expression (a + b + c + d)

a + b + c + d

= 6 + 2 + 6 + 6

= 20

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Answer:

kinetic energy body when it hits the ground is 98000 joule

1000joule = 1 kilojoule

, kinetic energy body when it hits the ground is 98  kilo-joule

Step-by-step explanation:

conservation of energy states that total energy of a system remains constant.

Potential energy of body = mgh

m = mass

g = gravitational pull = 9/8 m/s^2

h = height

kinetic energy = 1/2 mv^2

where v is the velocity of body

________________________________________

Total energy for this at any point is sum of potential energy and kinetic energy

total energy at height h

v= 0

PE = 250*9.8*40= 98,000

KE = 1/2 m0^2  = 0

total energy at when ball hits the ground

h=0

PE = 250*9.8*0 =

KE = 1/2 mv^2  

_______________________________________\

Applying conservation of energy

Total energy at height h = total energy at ground

98000 = KE

Thus, kinetic energy body when it hits the ground is 98000 joule

1000joule = 1 kilojoule

, kinetic energy body when it hits the ground is 98  kilo-joule

Answer:

Please find attached sketch of the path of the ball, having plot area and plot points, created with MS Excel

Step-by-step explanation:

Question;

The equation representing the path of the ball obtained from a similar question posted online are;

h₁ = -4·(t + 1)·(t - 5), h₂ = -4·(t - 2)² + 36, h₃ = -4··t² + 16·t + 20

The above equations represent the same path

The equation, h₁ = -4·(t + 1)·(t - 5), gives the roots of the height function, h(t), used in determining the height of the ball after time t

At (t + 1) = 0 (t = -1) or at (t - 5) = 0 (t = 5), the ball is at ground level

The ball reaches the ground, is at ground level at t = 1, and at t = 5 seconds after being tossed, where h(t) = 0

The equation of the path of the ball in vertex form, y = a·(x - 2)² + k, is h₂ = -4·(t - 2)² + 36, where, by comparison, we have;

The vertex of the ball = The maximum height reached by the ball  = (h, k) = (2, 36)

The coefficient of the quadratic term, t², is negative, therefore, the shape of the parabola is upside down, ∩, shape

The sketch of the path of the ball created with MS Excel, used in plotting the vertex, the initial value and the root points of the parabola, through which the ball passes and joining of the points with a 'smooth' curve is attached

Answer:

0.4

Step-by-step explanation:

1-0.6=0.4???????????????

The exponential function has the following form:

The function goes through the points ( 0, 19) and ( 2, 171 )

We will use the points to find the values of a and b

Substitute with the point ( 2, 171 ) and the value of a to find b

So, the answer will be:

The number of pairs that must be randomly selected from the set {1,3,5,7,9,11,13,15} to guarantee that at least one pair of these numbers add up to 16 is 5.

In the given set {1,3,5,7,9,11,13,15}, there are four pairs that sum up to 16, which are (1,15), (3,13), (5,11), and (7,9). Let k numbers be picked from the set. Consider k = 5, then using the pigeonhole principle, for one of the pairs described above, both of the endpoints must have been picked. Hence, if 5 numbers are picked from the set, then there exists a pair with a sum of 16.

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Answer:

x=18

Step-by-step explanation:

(3x-9)= 45 degrees because the angle you are solving for is half of a right angle (90 divided by 2 = 45)

so you take 45 and add 9 to it. That equals 54. To get the value of x, you divide 54 by 3 and you get 18.

hope this helps!

The two types of control charts for variables are the X-bar chart and the R-chart.

Control charts are widely used in statistical process control to monitor and analyze the variability and stability of processes over time. They are used for variables data, which are measurements or observations that can be quantified.

1. X-bar Chart: The X-bar chart is used to monitor the central tendency or average of a process. It tracks the sample means over time to detect any shifts or trends. It helps in assessing whether the process is in control or if there are any systematic variations from the target value.

2. R-chart (Range chart): The R-chart is used to monitor the variability or dispersion within a process. It tracks the ranges (the difference between the highest and lowest values) of samples over time. By analyzing the range values, it helps in understanding whether the process is stable and consistent or if there are any unusual variations or outliers.

Together, the X-bar chart and the R-chart provide valuable insights into process performance and allow for early detection of any deviations or abnormalities, helping organizations maintain quality standards and take appropriate corrective actions when necessary.

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Answer:

Step-by-step explanation:

A rectangle is shown. The rectangle will be enlarged by a scale factor of 1.5.

2 cm

5 cm

What will be the perimeter of the enlarged rectangle?

we find the two measures of the triangle

2 * 1.5 = 3 cm

5 * 1.5 = 7.5 cm

To find the perimeter of a rectangle, add the lengths of the rectangle's four sides

3 + 3 + 7.5 + 7.5 =

21 cm

If in March, Stinson Company completes its only two jobs in process, Jobs 10 and 11. The entry is Dr Finished goods inventory $50,000,Cr Work in process inventory $50,000.

Journal entry is used by companies to record their business transactions.

Preparation of the journal entries for the completion of the two jobs and the sale of Job 10

Mar. 31

Dr Finished goods inventory $50,000

(20,000+30,00)

Cr Work in process inventory $50,000

( To record the completion of the two jobs)

Mar. 31

Dr Cash $35,000

Cr Sales revenue $35,000

(To record the sale job 10)

Mar. 31

Dr Cost of goods sold $30,000

Cr Finished goods inventory $30,000

(To record the cost of the job sold)

Therefore the entry is Dr Finished goods inventory $50,000,Cr Work in process inventory $50,000.

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The complete question is:

In March, Stinson Company completes Jobs 10 and 11. Job 10 cost $20,000 and Job 11 $30,000. On March 31, Job 10 is sold to the customer for $35,000 in cash.Journalize the entries for the completion of the two jobs and the sale of Job 10.Date Account Titles and Explanation Debit CreditMar. 31 31 31

To provide Logical Proofs with line-by-line justifications for the following arguments,

Let's use the first 4 rules of inference.

Given below is the justification for each step of the proof with the applicable rule of Inference.

E > M1. A > (E > ~F) Premise2. H v (~F > M) Premise3. A Premise4. ~H  Premise5. A > E > ~F 1, Hypothetical syllogism6.

E > ~F 5,3 Modus Ponens 7 .

~F > M 2,3 Disjunctive Syllogism 8.

E > M 6,7 Hypothetical SyllogismIf

A is true, then E must be true because A > E > ~F.

Also, if ~H is true, then ~F must be true because H v (~F > M). And if ~F is true,

Then M must be true because ~F > M. Therefore, E > M is a valid  based on the given premises using the first four rules of inference.

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Answer:

x= 2/5

I suggest to always simplify if it's possible :)

Answer: Its 2097152

Step-by-step explanation:

Answer:

the length of the diagonal is 10cm

Step-by-step explanation:

C²=8²+6²

C²= 64+36

C²=100

√C²=√100

C=10cm

By showing that the corresponding sides are not proportional we know that the Triangles △fgh and △jih are not similar.

To prove that triangles △fgh and △jih are not similar, we need to show that at least one pair of corresponding sides is not proportional.
Let's compare the side lengths:

Side fg does not have a corresponding side in △jih.
Side gh in △fgh corresponds to side hi in △jih.
Side fh in △fgh corresponds to side ij in △jih.

By comparing the side lengths, we can see that side gh/hj and side fh/ij are not proportional.

Therefore, triangles △fgh and △jih are not similar.

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Triangle FGH (△FGH) is not similar to triangle JIH (△JIH) because their corresponding angles are not congruent and their corresponding sides are not proportional.

To prove that triangle FGH (△FGH) is not similar to triangle JIH (△JIH), we need to show that their corresponding angles and corresponding sides are not proportional.

1. Corresponding angles: In similar triangles, corresponding angles are congruent. If we compare the angles of △FGH and △JIH, we find that angle F in △FGH corresponds to angle J in △JIH, angle G corresponds to angle I, and angle H corresponds to angle H. Since the corresponding angles in both triangles are not congruent, we can conclude that the triangles are not similar.

2. Corresponding sides: In similar triangles, corresponding sides are proportional. Let's compare the sides of △FGH and △JIH. Side FG corresponds to side JI, side GH corresponds to side IH, and side FH corresponds to side HJ. If we measure the lengths of these sides, we can see that they are not proportional. Therefore, the triangles are not similar.

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Answer:

the value of X is 13..

Thank you ☺️

Answer:

the value of X is 13..

Step-by-step explanation: