Carla is a waitress at daybreak diner, and she earns $5 for each hour she works. Last week, she earned $148 total, including $68 in tips. How many hours did Carla work last week?

The second derivative with respect to t and evaluating it at t=0, we get the variance:

Var(X+Y) = Mx+y''(0) - [Mx+y'(0)]^2 = [-0.18(4e^-2t) + 0

Since X and Y are identically distributed, we can write the moment generating function of X as Mx(t) and that of Y as My(t).

Since X and Y are independent, the moment generating function of X + Y is given by the product of their individual moment generating functions:

Mx+y(t) = Mx(t)My(t)

We are given the moment generating function of X + Y as:

Mx+y(t) = 0.09e^-2t + 0.24e^t + 0.34 + 0.24e^t + 0.09e^2t

We can rewrite this as:

Mx+y(t) = 0.09(e^-2t + e^2t) + 0.48e^t + 0.34

Comparing this to the moment generating function of a normal distribution with mean 0 and variance σ^2, which is given by:

M(t) = e^(μt + σ^2t^2/2)

We see that the moment generating function of X + Y is that of a normal distribution with mean 0 and variance σ^2 = 1/2.

Thus, X + Y ~ N(0, 1/2).

Since X and Y are identically distributed, X ~ N(0, 1/4) and Y ~ N(0, 1/4).

Therefore,

P(X ≤ 0) = P(X - Y ≤ -Y) = P(Z ≤ -Y/√(1/2)),

where Z ~ N(0,1).

Since X and Y are identically distributed, we have

P(X - Y ≤ -Y) = P(Y - X ≤ X) = P(-Y + X ≤ X) = P(X ≤ Y)

So,

P(X ≤ 0) = P(X ≤ Y) = P(X - Y ≤ 0)

= P[(X+Y) - 2Y ≤ 0]

= P[Z ≤ 2(Y - X)/√2]

where Z ~ N(0,1).

Now, let's find the mean and variance of X + Y:

E[X + Y] = E[X] + E[Y] = 2E[X]

Since X and Y are identically distributed, we have E[X] = E[Y].

Thus, E[X + Y] = 2E[X] = 2E[Y]

And,

Var(X + Y) = Var(X) + Var(Y) = 2Var(X)

Since X and Y are identically distributed, we have Var(X) = Var(Y).

Thus, Var(X + Y) = 2Var(X)

Using the moment generating function of X + Y, we can find its mean and variance as follows:

Mx+y(t) = E[e^(t(X+Y))]

Taking the first derivative  with respect to t and evaluating it at t=0, we get the mean:

E[X+Y] = Mx+y'(0) = [0.09(-2e^-2t) + 0.48e^t + 0.24e^t + 0.18(2e^2t)]|t=0

= -0.18 + 0.24 + 0.18 = 0.24

Taking the second derivative with respect to t and evaluating it at t=0, we get the variance:

Var(X+Y) = Mx+y''(0) - [Mx+y'(0)]^2 = [-0.18(4e^-2t) + 0.

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Billy's health score go down  by 75% if his sugar resources are quadrupled

Let the amount of sugar in Billy's kitchen be denoted by S and the number of cookies he can bake be denoted by C. Let his health score be denoted by H. Then we have the following relationships:

C ∝ S (directly proportional)

C ∝ 1/H (inversely proportional)

Combining these two relationships, we get:

C ∝ S/H

If S is quadrupled, then C will also quadruple according to the first relationship. However, H will decrease by some percentage x according to the second relationship. To find x, we can use the fact that C is proportional to S/H:

C = k*S/H

where k is a constant of proportionality. If S is quadrupled, then C will also quadruple, so we have:

4C = k4S/H

C = kS/(H/4)

This tells us that if S is quadrupled, then C will be divided by H/4. In other words, C/H will be divided by 4. So, the percentage decrease in H can be found as follows:

C/H → (C/H)/4 = (S/H)/(4/k) → x = 100%*(1 - 1/4) = 75%

Therefore, if Billy's sugar resources are quadrupled, his health score will go down by 75%.

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We can say with 99% confidence that between 11% and 21% of adults have diabetes or pre-diabetes.

To find a 99% confidence interval for the proportion of adults with diabetes, we need to know the sample proportion and sample size. Let's assume that we have a random sample of n adults and p of them have diabetes. Then, the sample proportion is:

P = p/n

We can use the formula for the margin of error to calculate the range of plausible values for the true proportion of adults with diabetes:

margin of error = z*√(P(1-P)/n)

where z is the critical value from the standard normal distribution corresponding to a 99% confidence level. From a standard normal distribution table, we find that z = 2.576.

Using the formula for the margin of error, we can then calculate the lower and upper bounds of the confidence interval:

lower bound = P - margin of error

upper bound = P + margin of error

Rounding to the nearest whole percent, we get the final confidence interval.

For example, if our sample of n = 500 adults had 80 with diabetes, then the sample proportion would be:

P = 80/500 = 0.16

The margin of error would be:

margin of error = 2.576√(0.16(1-0.16)/500) = 0.045

The lower and upper bounds of the confidence interval would be:

lower bound = 0.16 - 0.045 = 0.115 (rounded to 11%)

upper bound = 0.16 + 0.045 = 0.205 (rounded to 21%)

Therefore, we can say with 99% confidence that between 11% and 21% of adults have diabetes or pre-diabetes.

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(a) Letting X1, ..., Xy be independent random variables and Using the union bound, we have P(|X1| + ... + |XN| ≥ t) ≤ P(|X1| ≥ t/N) + ... + P(|XN| ≥ t/N) ≤ 2N[tex]e^{(-tε/N)}[/tex] for all t > 0.

(b) Using the assumption that P(|Xi| < ∂) ≤ ∂ for some ∂ ∈ (0,1), we have P(Σ|Xi| < ∂N) ≥ 1 - NP(|Xi| ≥ ∂N) ≥ 1 - (1 - ∂)[tex]e^N[/tex].

Setting t = 2N[tex]e^ε[/tex], we obtain

which is equivalent to

By setting ∂ = 2Ne**ε/N, we get

P(Σ|Xi| < ∂) ≥ 1 - e**(-ε), and therefore,

Using the inequality (1 - x) ≤ e**(-x) for x > 0, we get (1 - ∂)**N ≤ e**(-N∂), and therefore, P(Σ|Xi| < ∂N) ≥ 1 - e**(-N∂) ≥ ∂**N.

Thus, we have shown that NP(Σ|Xi| < ∂N) ≥ ∂**N for some ∂ ∈ (0,1) and P(|X1| + ... + |XN| ≥ t) ≤ P(|X1| ≥ t/N) + ... + P(|XN| ≥ t/N) ≤ 2N[tex]e^{(-tε/N)}[/tex] for all t > 0

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The correct answer is c. Whenever the sample size is less than 5% of the population size, a finite population correction factor is needed in computing the standard deviation of the sampling distribution of sample means.

This correction factor takes into account the fact that when the sample size is small relative to the population, the variability of the sample means is affected. Without the correction factor, the standard deviation of the sampling distribution would be overestimated. However, if the sample size is large enough (more than 5% of the population size), the effect of finite population correction is negligible and can be ignored. If the population is infinite, the correction factor is not necessary as the sample size can be considered as a small proportion of the infinite population.

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The experimental probability of landing on heads is 12% greater than the theoretical probability of landing on heads. The correct option is C

Flipping a fair coin, the theoretical chance of landing on heads is 0.5, or 50%. The experimental probability is the ratio of the total number of coin flips to the number of times the coin landed on heads.

The experiment's experimental probability is 62/100 = 0.62 or 62% since the student flipped the coin 100 times and it came up heads 62 times.

We can see that by comparing the experimental and theoretical probabilities, 62% - 50% = 12%

So the experimental probability is 12% greater than the theoretical probability.

Therefore, The experimental probability of landing on heads is 12% greater than the theoretical probability of landing on heads.

Therefore, The correct option is C

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P(t < 1.94) ≈ 0.913 (rounded to three decimal places). For 14 degrees of freedom and P(-c < t < c) = 0.95, c ≈ 2.145

(a) To compute P(t < 1.94) for a t distribution with 3 degrees of freedom, you can use a t-distribution table or statistical software. Looking up the value in a table or using software, you will find that P(t < 1.94) ≈ 0.913.
(b) To find the value of c for a t distribution with 14 degrees of freedom such that P(-c < t < c) = 0.95, you can use a t-distribution table or statistical software again. For a 0.95 probability and 14 degrees of freedom, you will find that c ≈ 2.145.
So, the answers are:
(a) P(t < 1.94) ≈ 0.913 (rounded to three decimal places)
(b) For 14 degrees of freedom and P(-c < t < c) = 0.95, c ≈ 2.145

For the first part of the question, we need to use a t-distribution table or calculator to find the probability of the t variable being less than 1.94 with 3 degrees of freedom. Using a t-distribution table, we find that the probability is 0.950 with three decimal places. Therefore, P(t < 1.94) = 0.950.
For the second part of the question, we need to find the value of c such that the probability of the t variable being less than -c with 14 degrees of freedom is 0.025. Using a t-distribution table or calculator, we find that the value of c is 2.145 with three decimal places. Therefore, P(-c < t < c) = 0.95.

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Answer:

Alot

Step-by-step explanation:

Think abt it

Answer:

Step-by-step explanation:

If the squares of the two shorter sides add up to the square of the hypotenuse, the triangle contains a right angle.

[tex]\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ h=96 \end{cases}\implies V=\cfrac{\pi r^2 (96)}{3} \\\\[-0.35em] ~\dotfill\\\\ \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~\hspace{9em}\stackrel{\textit{since we know both Volumes are equal}}{\cfrac{4\pi r^3}{3}~~ = ~~\cfrac{\pi r^2 (96)}{3}}[/tex]

[tex]4\pi r^3=\pi r^2(96)\implies 4\pi r^2\cdot r=\pi r^2(96)\implies r=\cfrac{\pi r^2(96)}{4\pi r^2}\implies \boxed{r=24} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{ \textit{\LARGE cone} }{\cfrac{\pi (24)^2(96)}{3}}\implies \stackrel{ \textit{\LARGE sphere} }{\cfrac{4\pi (24)^3}{3}}\implies18432\pi ~~ \approx ~~ \text{\LARGE 57905.84}~units^3[/tex]

we can extend {1+x,1} to a basis of P3 by adding x^2.

To extend {1+x,1} to a basis of P3, we need to find one more polynomial that is linearly independent of these two. One way to do this is to choose a polynomial of degree 2, since we are working in P3. Let's try x^2.

We need to check if x^2 is linearly independent of {1+x,1}. This means we need to solve the equation a(1+x) + b(1) + c(x^2) = 0, where a, b, and c are constants.

Expanding this equation gives us a + ax + b + cx^2 = 0. Since x and x^2 are linearly independent, this means that a = 0 and c = 0. Therefore, we are left with just b(1) = 0, which means that b = 0 as well.

This shows that {1+x,1,x^2} is a linearly independent set, which means that it forms a basis of P3. Therefore, we have successfully extended {1+x,1} to a basis of P3 by adding x^2.

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Answer:

There seems to be some missing or incorrect information in the question. The given function f(x) = 1 5 1 4 x 2 is not well-formed and cannot be rewritten in the form f(x) = a(b)x. Please provide additional information or corrections to the question.

Step-by-step explanation:

1. The modal number of pairs of shoes owned by the children is 3.

2. The median number of pairs of shoes owned by the children is 3.

3. The Mean is 3.

1. The modal number of pairs of shoes owned by the children is 3.

2. The median number of pairs of shoes owned by the children

= 14/2 th term

= 7 th term

= 3

3. The Mean

= (1 x 2+ 2 x 3+ 3 x 5+ 4 x 2 + 5 x 1+ 6x 1)/ (2 +3 +5 +2 + 1 +1)

= 42/14

= 3

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The solutions to the equation to the nearest tenth are x = 10.1 and x = 2.9.

We have,

-2x² - 13x + 20 = -3x²

Combining like terms

-2x² - 13x + 20 = -3x²

x² - 13x + 20 = 0 (adding 3x² to both sides)

Now we can use the quadratic formula to solve for x:

x = (-b ± √(b² - 4ac)) / 2a

In this case,

a = 1, b = -13, and c = 20.

Substituting these values into the quadratic formula:

x = (-(-13) ± √((-13)² - 4(1)(20))) / 2(1)

x = (13 ± √(169 - 80)) / 2

x = (13 ± √(89)) / 2

So the solutions are:

x = (13 + √(89)) / 2

x ≈ 10.1

and

x = (13 - √(89)) / 2

x ≈ 2.9

Therefore,

The solutions to the equation to the nearest tenth are x ≈ 10.1 and x ≈ 2.9.

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The null hypothesis is H0: Median1 = Median2 = Median3 and the alternative hypothesis is Ha: Median1 ≠ Median2 ≠ Median3. The test statistic is H = 9.73. The p-value is 0.007. Reject H0. There is sufficient evidence to conclude that there is a significant difference in the time required by the three methods.

To determine whether there is a significant difference in the time required to complete a program evaluation with the three different methods, we will use an ANOVA test.

1. State the null hypothesis and alternative hypothesis:
H0: All populations of times are identical.
Ha: Not all populations of times are identical.

2. Find the value of the test statistic:
Using the given data, perform a one-way ANOVA test. You can use statistical software or a calculator with ANOVA capabilities to find the F-value (test statistic).

3. Find the p-value:
The same software or calculator used in step 2 will provide you with the p-value. Remember to round your answer to three decimal places.

4. State your conclusion:
Compare the p-value with the given significance level (α = 0.05).
- If the p-value is less than α, reject H0. There is sufficient evidence to conclude that there is a significant difference in the time required by the three methods.
- If the p-value is greater than or equal to α, do not reject H0. There is not sufficient evidence to conclude that there is a significant difference in the time required by the three methods.

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Answer:

Step-by-step explanation:

The height of the second cylinder should be 56.25% greater than the height of the first cylinder. (Option 1)

Let's assume the radius of the first cylinder to be 'r' and its height to be 'h'. So, its volume can be represented as V1 = πr^2h.

For the second cylinder, the radius of the base is 20% less than that of the first cylinder. So, the radius of the second cylinder can be represented as 0.8r. Let the height of the second cylinder be represented as 'H'. So, its volume can be represented as V2 = π(0.8r)²H.

As both cylinders have the same volume, we can equate the above two equations.

πr²h = π(0.8r)²H

h = (0.8)²H

H = (1/(0.8)²)h

H = (1.5625)h

Therefore, the height of the second cylinder should be 56.25% greater than the height of the first cylinder.

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Complete Question:

Two cylinders have the same volume, but the radius of the base of the second cylinder is 20% less than the radius of the base of the first. How much greater should be the height of the second cylinder be in comparison to the height in first?

Options:

In exercise 9.38(b), we are given a random sample Y1, Y2, ..., Yn from a normal distribution with mean μ and unknown variance σ^2. The likelihood function for this sample is:

L(μ, σ^2) = (2πσ^2)^(-n/2) exp[-∑(Yi-μ)^2/(2σ^2)]

To find the maximum likelihood estimator (MLE) of σ^2, we need to maximize the likelihood function with respect to σ^2 while holding μ constant. Taking the natural logarithm of the likelihood function and simplifying, we get:

ln L(μ, σ^2) = -n/2 ln(2π) - n/2 ln(σ^2) - ∑(Yi-μ)^2/(2σ^2)

Differentiating this expression with respect to σ^2 and setting the derivative equal to zero, we obtain:

d/dσ^2 ln L(μ, σ^2) = -n/(2σ^2) + ∑(Yi-μ)^2/(2σ^4) = 0

Solving for σ^2, we get:

σ^2 = ∑(Yi-μ)^2/n

Therefore, the MLE of σ^2 is the sample variance s^2 = ∑(Yi-ȳ)^2/(n-1), where ȳ is the sample mean. This is a well-known result in statistics and is based on the fact that the sample variance is an unbiased estimator of the population variance.

In conclusion, under the given conditions, the MLE of σ^2 is the sample variance s^2. This result is intuitive and makes sense since the sample variance is a natural estimator of the population variance based on the observed data. The normal distribution assumption is crucial for this result, as it allows us to derive the likelihood function and use maximum likelihood estimation to find the MLE of σ^2.

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The amount of time that students spend studying in the library in one sitting is normally distributed with a mean of 46 minutes and a standard deviation of 19 minutes. Hence the distribution of X is X ~ N (46, 19).

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To prove that df ^ dg = |df/dx df/dy| dxdy.|dg/dx dg/dy|, we can start by expanding the expression for the exterior product of the differentials df and dg.

df ^ dg = (df/dx dx + df/dy dy) ^ (dg/dx dx + dg/dy dy)

Using the distributive property of the exterior product, we can expand this expression as:

df ^ dg = (df/dx dx) ^ (dg/dx dx) + (df/dx dx) ^ (dg/dy dy) + (df/dy dy) ^ (dg/dx dx) + (df/dy dy) ^ (dg/dy dy)

Now, we can use the fact that the exterior product of two parallel vectors is zero, which means that (dx) ^ (dx) = (dy) ^ (dy) = 0. This allows us to simplify the expression as:

df ^ dg = (df/dx df/dy dy ^ dx) ^ (dg/dx dg/dy dy ^ dx)

Since dy ^ dx = -dx ^ dy, we can further simplify the expression as:

df ^ dg = -|df/dx df/dy| dx ^ dy ^ (dg/dx dg/dy) dx ^ dy

Now, we can use the fact that dx ^ dy = -dy ^ dx, which means that (dx ^ dy) ^ (dx ^ dy) = 0. This allows us to simplify the expression as:

df ^ dg = -|df/dx df/dy dg/dx dg/dy| (dx ^ dy) ^ (dx ^ dy)

Since (dx ^ dy) ^ (dx ^ dy) = 0, we can conclude that:

df ^ dg = 0

Therefore, we have proven that df ^ dg = |df/dx df/dy| dxdy.|dg/dx dg/dy|.

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The volume of the piñata that Mrs. Powell is making for her son's birthday, would be 2, 016 in ³

The piñata that Mrs. Powell is making, has a composite shape which means that you can find the volume by first finding the volume of the two composite shapes.

The volume of the cube is:

= Length x Width x Height

= 12 x 12 x 12

= 1, 728 in ³

Then the volume of the triangular prism :

= 1 / 2 x base x height x width

= 1 / 2 x 8 x 12 x 6

= 288 in ³

The volume of the pinata is:

= 1, 728 + 288

= 2, 016 in ³

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The efficient allocation of apples and bananas between the two individuals is:

A1 = B1 = 50 (allocated to individual 1)

A2 = 50 and B2 = 100 (allocated to individual 2)

What is utility?

In mathematics, utility refers to a measure of the preference or satisfaction an individual derives from consuming goods or services.

To recommend an efficient allocation of apples and bananas between the two individuals, we need to find a solution that maximizes the total utility of both individuals subject to the constraint that all the goods must be distributed. In other words, we need to solve the following optimization problem:

Maximize U1(A1, B1) + U2(A2, B2) subject to A1 + A2 = 100 and B1 + B2 = 150

Let's begin by solving for individual 1's optimal allocation. We can use the first-order conditions to find the optimal values of A1 and B1 that maximize U1(A1, B1). Taking partial derivatives with respect to A1 and B1 and setting them equal to zero, we get:

∂U1/∂A1 = 0 => 0 = 0

∂U1/∂B1 = 0 => 2 = 2B1/B2

Solving for B1/B2, we get B1/B2 = 1. This means that the optimal allocation for individual 1 is to receive an equal number of bananas and apples, i.e., A1 = B1 = 50.

Next, we solve for individual 2's optimal allocation. Following the same approach, we find that the optimal allocation for individual 2 is to receive all the remaining bananas and apples, i.e., A2 = 50 and B2 = 100.

Therefore, the efficient allocation of apples and bananas between the two individuals is:

A1 = B1 = 50 (allocated to individual 1)

A2 = 50 and B2 = 100 (allocated to individual 2)

This allocation is efficient because it maximizes the total utility of both individuals subject to the constraint that all the goods must be distributed. If we try to reallocate the goods in any other way, we will end up with a lower total utility for both individuals.

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The efficient allocation of apples and bananas between individuals 1 and 2 is as follows:

Individual 1 gets 60 apples and 75 bananas

Individual 2 gets 40 apples and 75 bananas

To determine the most efficient allocation of apples and bananas between individuals 1 and 2, we must maximize the total utility of both individuals while keeping in mind that all of the apples and bananas must be distributed.

From the constraint equation:

A1 + A2 = 100

B1 + B2 = 150

Now, let's write out the total utility function:

U = U1 + U2

U = 2A1 + B1 + 2A2 + B2 + 2A2B2

Using the Lagrangian method:

L = 2A1 + B1 + 2A2 + B2 + 2A2B2 - λ1(A1 + A2 - 100) - λ2(B1 + B2 - 150)

Taking the partial derivative of L with respect to each variable and equating them to zero, we get:

∂L/∂A1 = 2 - λ1 = 0

∂L/∂A2 = 2 + 4B2 - λ1 = 0

∂L/∂B1 = 1 - λ2 = 0

∂L/∂B2 = 1 + 2A2 - λ2 + 4A2B2 = 0

∂L/∂λ1 = A1 + A2 - 100 = 0

∂L/∂λ2 = B1 + B2 - 150 = 0

Solving these equations, we get:

λ1 = 2, λ2 = 1, A1 = 60, A2 = 40, B1 = 75, B2 = 75

Therefore, the efficient allocation of apples and bananas between individuals 1 and 2 is as follows:

Individual 1 gets 60 apples and 75 bananas

Individual 2 gets 40 apples and 75 bananas

This allocation maximizes the total utility of both individuals subject to the constraint that all the apples and bananas are distributed.

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From the trigonometric ratios, the value of sine trigonometric ratio for angle C, i.e., sin(C) in above right angled triangle CDE, is equals to the 0.55.

Trigonometry is a branch of mathematics. The trigonometric ratios are special measurements of a right triangle the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. The six trigonometric ratios in a right angled triangle are defined as sine, cosine, tangent, cosecant, secant, and cotangent. The symbols used for them are sin, cos, sec, tan, csc, cot. The three main ratios are defined as below

We have a right angled triangle CDE with 90° measure of angle D present in above figure. We have to determine the value of sine angle of C. Consider angle C priority,

Height or opposite of triangle = 11

Length of hypotenuse of triangle = 20

Using the above formula for sine trigonometric ratio, [tex]sin \: C = \frac{opposite}{hypotenuse}[/tex]

Substitute all known values in above formula, [tex]= \frac{11}{20}[/tex]

= 0.55

Hence, required value is 0.55.

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Complete question :

The above figure complete the question.

Find the value of sin C rounded to the nearest hundredth, if necessary.

20

E

11

D

Help PLEASE

10.-8.=2h

12:00-10:30=1.5h

2h-1.5h=0.5h

13km÷0.5h×2h=52km

Answer:

they jumped

Step-by-step explanation:

They jump because they want to escape Mario and Javier. Paula is very nervous because there are many people, it is not possible to escape quickly

I hope I’m right if not I’m sorry

The number of possibilities to make three-digit numbers from 1,4,5,6,3 that the first digit is even and the third digit is odd is 24.

1) To find the number of possibilities to make three-digit numbers from 1, 4, 5, 6, 3 where the first digit is even and the third digit is odd, follow these steps:

Identify the even numbers (for the first digit) - 4 and 6.
Identify the odd numbers (for the third digit) - 1, 3, and 5.
Calculate the possibilities for the second digit. Since we're using the remaining digits, there are 3 options left for each combination.
Multiply the possibilities together: 2 (even numbers) x 3 (second digit options) x 3 (odd numbers) = 18 possibilities.

2) To find the number of ways 5 students can seat in a circle, use the formula (n-1)!. Where n is the number of students.

For 5 students, there are (5-1)! = 4! = 4 x 3 x 2 x 1 = 24 ways for them to sit in a circle.

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Hi! It seems like your question might be about calculating the possible outcomes in an experiment. Based on the terms provided, I'll try my best to help you.

In an experiment with stages, the possible outcomes can be calculated using the multiplication principle. If there are two possible outcomes in the first stage and three possible outcomes in the second stage, you can multiply these numbers to find the total possible outcomes.

Total outcomes = (Outcomes in stage 1) x (Outcomes in stage 2)

Total outcomes = 2 x 3 = 6

Based on the given options, none of them match the calculated total outcomes.

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It is true that, If (a, b) = 1 then if a and b are odd numbers, then (a + b, a − b) = 2. Otherwise, (a + b, a − b) = 1

GCD, or Greatest Common Divisor, is a fundamental concept in number theory. It is defined as the largest positive integer that divides both two or more integers without leaving a remainder.

In other words, the GCD of two numbers is the largest positive integer that divides both of them evenly.

Here we have

Let's consider two cases:

Case 1: a and b are odd numbers

In this case, we can express a and b as:

a = 2k+1

b = 2m+1

where k and m are integers.

Then,

a+b = (2k+1) + (2m+1) = 2(k+m+1)

a-b = (2k+1) - (2m+1) = 2(k-m)

We can see that both a+b and a-b are even.

Therefore, (a+b, a-b) is at least 2.

Now, let's show that (a+b, a-b) cannot be larger than 2:

Suppose, for contradiction, that (a+b, a-b) = d > 2.

Then, d divides both (a+b) and (a-b).

We can write (a+b) and (a-b) as:

=> a+b = dx

=> a-b = dy

where x and y are integers.

Adding the above two equations, we get:

2a = d(x+y)

Since a is odd, d must be odd as well.

Substituting for 'a' in terms of x and y, we get:

=> 2(2k+1) = d(x+y)

=> 4k+2 = d(x+y)

=> 2(2k+1) = 2d(x+y)/2

=> 2k+1 = d(x+y)/2

We can see that d must divide 2k+1 since x and y are integers.

However, we know that (a,b) = 1, which means that a and b do not have any common factors other than 1.

Since a is odd, 2 does not divide a.

Therefore, d cannot be greater than 2, which is a contradiction.

Hence,

(a+b, a-b) = 2 when a and b are odd numbers.

Case 2: a and b are not both odd numbers

Without loss of generality,

Let's assume that a is even and b is odd.

Then, a+b and a-b are both odd.

Since odd numbers do not have any factors of 2, (a+b, a-b) = 1.

Therefore,

(a+b, a-b) = 2 if a and b are both odd and (a+b, a-b) = 1 if a and b are not both odd.

By the above explanation,

It is true that, If (a, b) = 1 then if a and b are odd numbers, then (a + b, a − b) = 2. Otherwise, (a + b, a − b) = 1

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a. If the score recorded in the grade book is the total number of points earned on the two parts,  the expected recorded score e(x 1 y) is 11.6.

b.  If the maximum of the two scores is recorded, the expected recorded score 10.08.

a) The expected recorded score is given by:

e(x + y) = ΣΣ(x + y) * p(x, y)

So, we have:

e(x + y) = (0+0)*0.02 + (5+0)*0.04 + (10+0)*0.06 + (15+0)*0.02 + (5+10)*0.15 + (10+10)*0.20 + (15+10)*0.15 + (5+15)*0.01 + (10+15)*0.14 + (15+15)*0.01

Simplifying:

e(x + y) = 0.02(0 + 0) + 0.04(5 + 0) + 0.06(10 + 0) + 0.02(15 + 0) + 0.15(5 + 10) + 0.20(10 + 10) + 0.15(15 + 10) + 0.01(5 + 15) + 0.14(10 + 15) + 0.01(15 + 15)

e(x + y) = 11.6

So, the expected recorded score is 11.6.

b) The expected recorded score if the maximum of the two scores is recorded is given by:

e(max(x, y)) = ΣΣ(max(x, y)) * p(x, y)

So, we have:

e(max(x, y)) = max(0, 5)*0.06 + max(5, 0)*0.04 + max(10, 0)*0.06 + max(15, 0)*0.02 + max(5, 10)*0.15 + max(10, 10)*0.20 + max(15, 10)*0.15 + max(5, 15)*0.01 + max(10, 15)*0.14 + max(15, 15)*0.01

Simplifying:

e(max(x, y)) = 0.065 + 0.045 + 0.0610 + 0.0215 + 0.1510 + 0.2010 + 0.1515 + 0.0115 + 0.1415 + 0.0115

e(max(x, y)) = 10.08

So, the expected recorded score is 10.08.

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