Show the enteiws to close $2000 in expense, $5000 in revenue, and
$500 in dividens.

The probability that all 11 tests will be positive is 0.8007

The probability that at least one test will be negative is 0.1993

From the question, we have the following parameters that can be used in our computation:

p = 0.98

n = 11

The probability that all 11 tests will be positive is

P = pⁿ

So, we have

P = 0.98¹¹

Evaluate

P = 0.8007

Here, we use

P = 1 - P(No negative)

So, we have

P = 1 - 0.8007

Evaluate

P = 0.1993

Hence, the probability is 0.1993

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Answer:

24

Brainliest?

Answer:

nxbxbxxbznznxnxnxnxnxnxnxbxbbxbnxxnxnx

Step-by-step explanation:

xnncxbbxbxbxbxbxbx hahahahahahahahahahahah

The values of c₁ and c₂ that satisfy the initial values u(0) = 60 and u'(0) = 56 are:

c₁ = 148 / (3e²)

c₂ = (20 - 148/9)e⁴

The given solution, u(t) = c₁e² + c₂e⁻⁴, indeed satisfies the given ordinary differential equation (ODE) u"(t) + u'(t) - 12u(t) = 0. To find the values of c₁ and c₂ such that the solution satisfies the initial values u(0) = 60 and u'(0) = 56, we substitute these values into the solution.

First, let's find u(0) by substituting t = 0 into the solution:

u(0) = c₁e² + c₂e⁻⁴

Since u(0) = 60, we have:

60 = c₁e² + c₂e⁻⁴    (Equation 2)

Next, let's find u'(0) by differentiating the solution with respect to t and substituting t = 0:

u'(t) = 2c₁e² - 4c₂e⁻⁴

u'(0) = 2c₁e² - 4c₂e⁻⁴

Since u'(0) = 56, we have:

56 = 2c₁e² - 4c₂e⁻⁴    (Equation 3)

Now we have a system of two equations (Equations 2 and 3) with two unknowns (c₁ and c₂). We can solve this system to find the values of c₁ and c₂.

To do that, let's first divide Equation 3 by 2:

28 = c₁e² - 2c₂e⁻⁴

Next, let's multiply Equation 2 by 2:

120 = 2c₁e² + 2c₂e⁻⁴

Adding the two equations, we get:

148 = 3c₁e²

Dividing both sides by 3e², we find:

c₁ = 148 / (3e²)

Substituting this value of c₁ back into Equation 2, we can solve for c₂:

60 = (148 / (3e²))e² + c₂e⁻⁴

60 = 148/3 + c₂e⁻⁴

60 - 148/3 = c₂e⁻⁴

20 - 148/9 = c₂e⁻⁴

c₂ = (20 - 148/9)e⁴

Therefore, the values of c₁ and c₂ that satisfy the initial values u(0) = 60 and u'(0) = 56 are:

c₁ = 148 / (3e²)

c₂ = (20 - 148/9)e⁴

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The joint probability density function (pdf) is f(x,y) = 0.03 for 0 < x < 1 and 0 < y < 3.

The joint pdf, f(x,y), represents the probability density function for the random variables X and Y. Given that X is uniformly distributed on (0,10), we have fX(x) = 0.1 for 0 < x < 10. The probability of X being less than 1 is 1/10, so the conditional pdf f(x|X<1) = 0.1 for 0 < x < 1.

Furthermore, Y is uniformly distributed on (0,3), so fY(y) = 0.1 for 0 < y < 3. To find the joint pdf, we multiply the conditional pdf of X with the marginal pdf of Y: f(x,y) = f(x|X<1) * fY(y) = 0.1 * 0.1 = 0.01 for 0 < x < 1 and 0 < y < 3. Therefore, the joint pdf is f(x,y) = 0.01 for 0 < x < 1 and 0 < y < 3.

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Answer:

22cm

Step-by-step explanation:

Length of arc = [tex]\frac{theta}{360}[/tex] * 2[tex]\pi[/tex]r

                      = [tex]\frac{45}{360}[/tex] * 2 * [tex]\frac{22}{7}[/tex] * 28

                      = 22 cm

Answer:

a regular person would gave moved 4 to 5 times

The most probable values for X and Y are X = 11.75 and Y = 1.1, respectively.

To solve the system of equations using the tabular or matrix method, we first convert the given equations into matrix form. We create a coefficient matrix A by arranging the coefficients of the variables X and Y, and a constant vector B by placing the constants on the other side of the equations.

To solve the system of equations using the tabular method or matrix method, we'll first write the equations in matrix form. Let's define the coefficient matrix A and the constant vector B:

A = | 1 2 |

| 2 -3 |

| 2 -1 |

B = | 10.5 |

| 5.5 |

| 10.0 |

Now, we can solve the system of equations by finding the inverse of matrix A and multiplying it with vector B:

[tex]A^{(-1)[/tex] = | 1.5 1 |

| 0.4 0.2 |

X = [tex]A^{(-1)[/tex] * B

Multiplying [tex]A^{(-1)[/tex] with B, we get:

X = | 1.5 1 | * | 10.5 | = | 11.75 |

| 0.4 0.2 | | 5.5 | | 1.1 |

Therefore, the most probable values for X and Y are X = 11.75 and Y = 1.1, respectively.

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Answer:

in right angled triangleBCD

BC=√{DC²-BC²)=√{10²-6²)=8

again in right angled triangle ABC

AB=√(BC²-AC²)

x=√(8²-7²)=3.87

Answer:

the mean and the standard deviation is 0.92 and 0.01808 respectively

Step-by-step explanation:

The computation of the mean and the standard deviation is shown below:

The mean is 0.92

And, the standard deviation is

= √0.92 × (1 - 0.92) ÷ √225

= √0.92 × 0.08 ÷ √225

= √0.0736 ÷ √225

= √3.27

= 0.01808

Hence, the mean and the standard deviation is 0.92 and 0.01808 respectively

Answer:

c - 13,600ft^2

Step-by-step explanation:

200 x 80 = 16,000ft^2

40 x 60 = 2,400ft^2

16,000ft^2 - 2,400ft^2 = 13,600ft^2

The solution to the system of inequalities is the region above the x-axis and to the right of the line y = 1, shaded in green.

Given system of inequalities is: x - 2y ≤ 3 ...(1)3x + 4y ≥ 12 ...(2)x ≥ 0 ...(3)y ≥ 1 ...(4)

We graph the lines x - 2y = 3 and 3x + 4y = 12 and shade the appropriate regions.

Let's start with the line x - 2y = 3.

We rewrite this as y = (1/2)x - 3/2 and plot the line as shown below: graph{(1/2)x - 3/2 [-10, 10, -5, 5]}

Now we determine which side of the line we want to shade.

Since the inequality is of the form ≤, we shade below the line y = (1/2)x - 3/2 (including the line itself) as shown below: graph {(1/2)x - 3/2 [-10, 10, -5, 5](-10,-5)--(10,0)}

Next, we graph the line 3x + 4y = 12. We rewrite this as y = (-3/4)x + 3 and plot the line as shown below: graph{(-3/4)x + 3 [-10, 10, -5, 5]}

We determine which side of the line we want to shade. Since the inequality is of the form ≥, we shade above the line y = (-3/4)x + 3 (including the line itself) as shown below: graph{(-3/4)x + 3 [-10, 10, -5, 5](-10,4)--(10,0)}

Finally, we shade the region that satisfies x ≥ 0 and y ≥ 1.

This is the region above the x-axis and to the right of the line y = 1 as shown below: graph{(-3/4)x + 3 [-10, 10, -5, 5](-10,4)--(10,0)(0,1)--(10,1)[above]}

The shaded region is the region that satisfies all three inequalities.

Thus, the solution to the system of inequalities is the region above the x-axis and to the right of the line y = 1, shaded in green.

We graph the lines x - 2y = 3 and 3x + 4y = 12 and shade the appropriate regions.

Let's start with the line x - 2y = 3. We rewrite this as y = (1/2)x - 3/2 and plot the line.

Now we determine which side of the line we want to shade. Since the inequality is of the form ≤, we shade below the line y = (1/2)x - 3/2 (including the line itself).

Next, we graph the line 3x + 4y = 12. We rewrite this as y = (-3/4)x + 3 and plot the line. We determine which side of the line we want to shade.

Since the inequality is of the form ≥, we shade above the line y = (-3/4)x + 3 (including the line itself).

Finally, we shade the region that satisfies x ≥ 0 and y ≥ 1.

This is the region above the x-axis and to the right of the line y = 1. The shaded region is the region that satisfies all three inequalities.

Thus, the solution to the system of inequalities is the region above the x-axis and to the right of the line y = 1, shaded in green.

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The "margin-of-error" is 3.6 and the "sample-mean" is 18.7 based on the given confidence interval (15.1, 22.3).

To find the "margin-of-error" and the "sample-mean" from the given confidence interval, we use the formula:

Confidence-Interval = Sample mean ± Margin of error,

In this case, the given confidence-interval is (15.1, 22.3),

To find the margin-of-error, we need to consider the range between the upper and lower bounds of the confidence interval and divide it by 2,

The "Margin-of-error" is = (Upper bound - Lower bound)/2,

"Margin-of-error" is = (22.3 - 15.1)/2 = 3.6,

So, the margin of error is 3.6.

To find "sample-mean", we calculate average of the upper and lower bounds of the confidence-interval,

The "Sample-Mean" is = (Upper bound + Lower bound)/2,

"Sample-Mean" is = (22.3 + 15.1)/2 = 18.7,

Therefore, the "sample-mean" is 18.7.

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The given question is incomplete, the complete question is

Use the confidence interval to find the margin of error and the sample mean. (15.1, 22.3).

In Hausdorff-space "X", if A and B are disjoint "compact-subspaces", then there is disjoint "open-sets" U and V such that A is contained in U and B is contained in V, this is because by Hausdorff-Property, the existence of disjoint open neighborhoods for any two "distinct-points".

To prove the existence of disjoint "open-sets" U and V with A⊂U and B⊂V, where A and B are "compact-subspaces" of "Hausdorff-space" X,

Step (1) : A and B are disjoint compact-subspaces, we use Hausdorff property to find "open-sets" Uₐ and [tex]U_{b}[/tex] such that "A⊂Uₐ" and "B⊂[tex]U_{b}[/tex]", and "Uₐ∩[tex]U_{b}[/tex] = ∅". This can be done for every pair of points in A and B, respectively, because X is Hausdorff.

Step (2) : We consider, set U = ⋃ Uₐ, where "union" is taken over all of Uₐ for each-point in A. U is = union of "open-sets", hence open.

Step (3) : We consider set V = ⋃ [tex]U_{b}[/tex], where union is taken over for all [tex]U_{b}[/tex] for "every-point" in B. V is also a union of open-sets and so, open.

Step (4) : We claim that U and V are disjoint. Suppose there exists a point x in U∩V. Then x must be in Uₐ for some point a in A and also in [tex]U_{b}[/tex] for some point b in B. Since A and B are disjoint, a and b are different points. However, this contradicts the fact that Uₐ and [tex]U_{b}[/tex] are disjoint open sets.

Therefore, U and V are disjoint open sets with A⊂U and B⊂V.

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The given question is incomplete, the complete question is

Let A and B be disjoint compact subspaces of a Hausdorff space X. Show that there exist disjoint open sets U and V, with A⊂U and B⊂V.

Answer:

7

Step-by-step explanation:

the answer I got was 7 hope this helped

Answer:

True

Step-by-step explanation:

The wording of the question is a little tricky, but here's what I think.

If a function f(x) is not defined at a point c, then the function has a discontinuity at that point. In order for a function to be continuous on an interval, it must be defined and have no abrupt changes or jumps within that interval. Since f(x) is not defined at c, it violates the condition of continuity, and therefore f(x) cannot be continuous on any interval that includes c.

The given statement "If f(x) is not defined at c, then f(x) cannot be continuous on any interval." is false because it does not automatically mean that f(x) cannot be continuous on any interval.

Continuity of a function depends on the behavior of the function around the point of interest, rather than just the absence of a definition at a single point. A function can still be continuous on an interval except at the specific point where it is not defined.

For example, consider the function f(x) = 1/x. This function is not defined at x = 0, but it is continuous on any interval that does not include x = 0. This is because f(x) approaches positive or negative infinity as x approaches 0 from the left or right side, respectively, indicating that there is no abrupt jump or discontinuity.

In general, the continuity of a function is determined by its behavior around a point, including its limit as x approaches that point. The absence of a definition at a single point does not automatically imply that the function cannot be continuous on any interval.

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The correct statement regarding r squared is:

r squared equals 0.4268 meaning that Years of Experience explains 42.68 percent of the variability in Salary.

Part A: The correct null and alternate hypotheses are:

H₀: β₁=0;

H₁: β₁≠0.

Part B: The correct p-value for the test in Q1 is 1.74e-13.

Part C: The fitted regression model from this output in Q1 is:

E(Salary) = 31.8387 + 2.8205 x Years of Experience.

Part D: The correct statement regarding r squared is:

r squared equals 0.4268 meaning that Years of Experience explains 42.68 percent of the variability in Salary.

Explanation: The output shows a multiple linear regression model:

Salary=β0+β1x

Years of Experience + ϵ.β0 is the intercept and represents the expected mean salary for an individual with 0 years of experience.

β1 is the slope and represents the expected change in salary due to one year increase in experience.

ϵ is the error term (deviation from the expected salary).

The correct null and alternate hypotheses are:

H₀: β₁=0 (there is no linear relationship between salary and years of experience).

H₁: β₁≠0 (there is a linear relationship between salary and years of experience).

The correct p-value for the test in Q1 is 1.74e-13, which is much smaller than the significance level of 0.05.

Thus, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that there is a linear relationship between salary and years of experience.

The fitted regression model from this output in Q1 is:

E(Salary) = 31.8387 + 2.8205 x Years of Experience.

The coefficient of determination, or R-squared, is a statistical measure that shows how well the regression model fits the observed data.

It is the proportion of the variance in the dependent variable that is explained by the independent variable(s).

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Answer:

-6

Step-by-step explanation:

Step One: The goal is to isolate the x. So first, we would do 4-22, which is -18. The equation is: 3x=-18.

Step Two: Lastly, we need to divide by three to completely isolate the x. This is -6.

Answer:

78

Step-by-step explanation:

Answer:

1440 and 144 on edg 2020-2021

Step-by-step explanation:

Answer:

(x-4)^2 -4 = y

Step-by-step explanation:

The graph is translated 4 to the right and 4 downwards.

Answer:

17cm

Step-by-step explanation:

The two equal sides are 14cm long so 45 - 14 - 14 = 17

Answer:

6/9

Step-by-step explanation:

The answer is A, 3.9a + b

Step-by-step explanation:

youre welcome

Answer:

b

Step-by-step explanation:

oi did the quiz f. 6373737

The P(you win) = 1/16P(you lose) = 15/16 and Average winnings, µ, = -5/16.

The probability of winning (P) and the probability of losing (Q) are both possible outcomes from a coin toss experiment. The probability of winning is given by the formula

P = Number of ways to win/ Total number of possible outcomes.The probability of losing is given by the formula

Q = Number of ways to lose/Total number of possible outcomes.

Formulae used to calculate probability are:

P = Number of ways to win/ Total number of possible outcomes

P = Number of outcomes in which all four tosses are heads/ Total number of possible outcomes

When a coin is tossed four times, the total number of possible outcomes is 2 × 2 × 2 × 2 = 16.

P = Number of outcomes in which all four tosses are heads/ Total number of possible outcomes

P = 1/16P (you win) = 1/16

The probability of losing is given by the formula

Q = Number of ways to lose/Total number of possible outcomes.

Q = 15/16P (you lose) = 15/16

Average winnings, µ, can be calculated as follows:

Let's say X represents the amount of money you win. When you win, you get $10, and when you lose, you lose $1.

X = -1 when you loseX = 10 when you winUsing the formula

µ = ∑ (X × P), we can calculate the average winnings,

µ = (-1 × 15/16) + (10 × 1/16)µ = -15/16 + 10/16µ = -5/16.

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Given information: A coin is tossed four times. You bet $1 that heads will come up on all four tosses. If this happens, you win $10. Otherwise, you lose your $1 bet.

Therefore, the answers are:

P(you win) = 0.0625

P(you lose) = 0.9375

Average winnings, µ, = -$0.31

Let A be the event of getting heads on the four tosses. Since the coin is tossed four times, the possible outcomes are 2^4 = 16. Thus the probability of getting heads on each toss is 0.5. Therefore, the probability of getting heads on all four tosses is:

P(A) = (0.5)^4

= 0.0625

Let B be the event of not getting heads on all four tosses. Thus:

B = 1 − P(A)

= 1 − 0.0625

= 0.9375

The winning amount for getting all four heads is $10, and the losing amount is $1. Thus, the average winnings is:

µ = (10 × 0.0625) − (1 × 0.9375)

µ = 0.625 − 0.9375

µ = -0.3125

Therefore, the answers are:

P(you win) = 0.0625

P(you lose) = 0.9375

Average winnings, µ, = -$0.31

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Answer:

40,000

Step by step explanation:

842,963

840,000

40,000

If a and n are relatively prime (gcd(a, n) = 1), it does not guarantee that the equation x ≡ a (mod n) has solutions.

If a and n are relatively prime, denoted by gcd(a, n) = 1, it means that a and n do not have any common factors other than 1. However, this does not guarantee that the equation x ≡ a (mod n) has solutions.

The equation x ≡ a (mod n) represents a congruence relation, where x is congruent to a modulo n. This equation implies that x and a have the same remainder when divided by n.

To have solutions for this congruence equation, it is necessary for a to be congruent to some number modulo n. In other words, a must lie in the residue classes modulo n. However, the fact that gcd(a, n) = 1 does not ensure that a is congruent to any residue modulo n, hence not guaranteeing the existence of solutions for the equation.

Therefore, when n = 1, we cannot conclude that the equation x ≡ a (mod n) has solutions.

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Answer:

C

Step-by-step explanation:

C

The answer is x = 15.

15 or 20.33 are the possible values of the x.

Any mathematical statement that includes numbers, variables, and an arithmetic operation between them is known as an expression or algebraic expression. In the phrase 4m + 5, for instance, the terms 4m and 5 are separated from the variable m by the arithmetic sign +.

We know that sin(x+37)°=cos(90°-(x+37)°) and cos(2x+8)°=sin(90°-(2x+8)°)

So we have sin(x+37)°=cos(2x+8)° becomes sin(x+37)°=sin(82°-2x)

For the above equation to be true, either of the following must hold:

x+37 = 82 - 2x (since the sin function is periodic)

x+37 = 180 - (82-2x)

Solving the first equation for x gives:

3x = 45

x = 15

Solving the second equation for x gives:

3x = 61

x = 20.33 (rounded to two decimal places)

Therefore, the possible values of x are 15 or 20.33 (rounded to two decimal places).

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Answer:

FGH DEJ .................