Alma makes `5` cups of her favorite shade of purple paint by mixing `3` cups of blue paint, `1\frac{1}{2}` cups of red paint, and `\frac{1}{2}` a cup of white paint.



Alma has `2` cups of white paint.



How much blue paint and red paint will Alma need to use with the `2` cups of white paint?

Answer: 6 turns

Step-by-step explanation:

The slope and the y intercepts of the given graph which shows height of the candlesticks over time is :

Slope = -2/3

Y intercept = 9

The graph is that of the height of the candlesticks over time.

The line passes through two points (0, 9) and (3, 7).

Slope of the line can be calculated as,

Slope = (7 - 9) / (3 - 0) = -2/3

Hence the slope id negative two thirds.

y intercept of a graph is the y coordinate of the point where the line touches the Y axis.

The x coordinate will be 0 there.

The line passes through (0, 9).

So y intercept = 9

Hence the slope and the y intercept are -2/3 and 9 respectively.

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So the vector form of the quadratic function y = x² - 6x + 7 is: y = (x - 3)² - 2.

To change from standard form to vertex form, we need to complete the square.

First, we group the x-terms together and factor out any common coefficient of x², giving:

y = x² - 6x + 7

y = 1(x² - 6x) + 7

Next, we need to add and subtract a constant inside the parentheses to complete the square. To determine this constant, we take half of the coefficient of x (-6) and square it:

(-6/2)² = 9

So we add and subtract 9 inside the parentheses:

y = 1(x² - 6x + 9 - 9) + 7

Now we can factor the quadratic expression inside the parentheses as a perfect square:

y = 1[(x - 3)² - 9] + 7

Simplifying and rearranging terms, we get:

y = (x - 3)² - 2

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The 90% CI for the change in blood pressure is (-5.843, 7.977)., the 99.9% CI for the change in blood pressure is (-14.077, 16.211).

a) To calculate a 90% confidence interval (CI) for the change in blood pressure, we use the formula:

CI = x ± t(α/2, df) * (s/√n)

where x is the sample mean of the change in blood pressure, t(α/2, df) is the t-value for the desired confidence level and degrees of freedom, s is the sample standard deviation of the change in blood pressure, and n is the sample size.

Using the given data, we have:

x = 1.067
s = 26.691
n = 15
df = n - 1 = 14

From the t-distribution table, the t-value for a 90% confidence level and 14 degrees of freedom is 1.761.

Plugging in the values, we get:

CI = 1.067 ± 1.761 * (26.691/√15) = (-5.843, 7.977)

Therefore, the 90% CI for the change in blood pressure is (-5.843, 7.977).

b) To calculate a 99.9% CI for the change in blood pressure, we use the same formula but with a different t-value:

t(α/2, df) = 3.922 for a 99.9% confidence level and 14 degrees of freedom.

Plugging in the values, we get:

CI = 1.067 ± 3.922 * (26.691/√15) = (-14.077, 16.211)

Therefore, the 99.9% CI for the change in blood pressure is (-14.077, 16.211).

c) The interval calculated in part (a) does include 0, while the interval calculated in part (b) does not include 0. This is important because if the interval includes 0, it means we cannot conclude that there is a significant difference in blood pressure before and after the medication. On the other hand, if the interval does not include 0, it means we can be confident that there is a significant difference.

d) To conduct a one sample t-test with p = 0 and a = 10, we first calculate the t-statistic using the formula:

t = (x - p) / (s/√n)

where x is the sample mean of the change in blood pressure, s is the sample standard deviation of the change in blood pressure, and n is the sample size.

Plugging in the values, we get:

t = (1.067 - 0) / (26.691/√15) = 0.444

From the t-distribution table, the t-value for a one-tailed test with 14 degrees of freedom and a significance level of 10% is 1.345.

Since our calculated t-value (0.444) is less than the critical t-value (1.345), we fail to reject the null hypothesis that there is no significant difference in blood pressure before and after the medication. Therefore, the results are not consistent with the 90% CI calculated in part (a).

e) To conduct a one sample t-test with y = 0 and a = 0.001, we use the same formula as in part (d):

t = (x - y) / (s/√n)

Plugging in the values, we get:

t = (1.067 - 0) / (26.691/√15) = 0.444

From the t-distribution table, the t-value for a one-tailed test with 14 degrees of freedom and a significance level of 0.001 is 3.746.

Since our calculated t-value (0.444) is less than the critical t-value (3.746), we fail to reject the null hypothesis that there is no significant difference in blood pressure before and after the medication. Therefore, the results are not consistent with the 99.9% CI calculated in part (b).

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The answer is 64.  To find the number of homozygous dominant individuals in the population.

We need to use the Hardy-Weinberg equation:

p^2 + 2pq + q^2 = 1

Where:
p = frequency of dominant allele
q = frequency of recessive allele

Since the dominant allele frequency is 0.8, we can assume that the recessive allele frequency is 0.2 (since p + q = 1).

To find the frequency of homozygous dominant individuals (p^2), we simply square the frequency of the dominant allele:

p^2 = (0.8)^2 = 0.64

To find the number of homozygous dominant individuals, we multiply the frequency by the total population size:

0.64 x 100 = 64

Therefore, there are 64 homozygous dominant individuals in the population.

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We need to use the exponential distribution formula and the concept of the sum of exponential variables. Let X be the time between two consecutive breakdowns and Y be the time to repair a broken plane.

Both X and Y follow an exponential distribution with mean 1/2 and 1/7, respectively.
The time T until a plane is available again after a breakdown can be expressed as T=X+Y. The distribution of T is the convolution of the distributions of X and Y, which is also an exponential distribution with mean 9/14.
Let Z be the number of planes available after all breakdowns have occurred and repairs have been made. Z is a binomial variable with parameters n=100 and p=P(T>1), where P(T>1) is the probability that a breakdown occurs and is repaired in more than one week.
We want to find the minimum number of repairmen, denoted by k, such that P(Z≥90)≥0.95. Using the complement rule, we can rewrite this as P(Z<90)≤0.05.
Since Z follows a binomial distribution, we can use the normal approximation to compute P(Z<90). The mean of Z is np=100p, and the variance of Z is np(1-p). Therefore, Z can be approximated by a normal distribution with mean μ=100p and standard deviation σ=sqrt(np(1-p)).
To find p, we can use the fact that P(T>1)=exp(-λT), where λ=1/mean(X+Y)=14/9. Thus, p=1-exp(-λ). Plugging in the values, we get p=0.432.
Now, we can standardize Z by subtracting its mean and dividing by its standard deviation: Z'=(Z-μ)/σ. Then, we can use a standard normal table or calculator to find the corresponding probability: P(Z'<(90-μ)/σ).
Finally, we can solve for k using the inverse standard normal function: (90-μ)/σ=invNorm(0.05). Plugging in the values, we get k=30.
Therefore, we need at least 30 repairmen to ensure that there is at least a 95% chance that 90 or more planes are available.

To determine the number of repairmen needed for Allbest Airlines to ensure there is a 95% chance that 90 or more planes are available, we'll use the following terms: airlines, breakdowns, and exponential.
Step 1: Determine the required probability.
Since we want at least 90 planes available, it means that no more than 10 planes can be under repair at a time. Thus, we need to calculate the probability that 10 or fewer planes are under repair.
Step 2: Calculate the probability of a single plane being under repair.
Planes break down twice a year and take one week to fix. Therefore, the probability of a single plane being under repair in any given week is 2 breakdowns per year * 1 week per breakdown / 52 weeks per year = 1/26.
Step 3: Use the exponential distribution.
We can model the probability of a certain number of planes being under repair using the exponential distribution.
Step 4: Calculate the probability of 10 or fewer planes being under repair.
Using a cumulative distribution function (CDF) for the exponential distribution, we can calculate the probability of 10 or fewer planes being under repair to be at least 95%. This can be done using statistical software or a calculator with the appropriate functionality.
Step 5: Determine the number of repairmen needed.
If the probability calculated in step 4 is 95% or higher, the current number of repairmen is sufficient. If not, we need to increase the number of repairmen until the probability is at least 95%. This can be done by reducing the average repair time per plane, which in turn reduces the probability of a plane being under repair at any given time.
In conclusion, to ensure that there is a 95% chance that 90 or more planes are available for Allbest Airlines, we need to determine the number of repairmen needed based on the exponential distribution and probability calculations as outlined in the steps above.

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The largest angle is B and the smallest angle is C

When we use the cosine rule, we can see that;

c^2 = a^2 + b^2 - 2ab Cos C

4.5^2 = 6.5^2 + 8.5^2 - 2(6.5 * 8.5) CosC

20.25 = 114.5 - 110.5CosC

20.25 - 114.5 = - 110.5CosC

CosC = 0.8529

C =Cos-1  0.8529

C = 31 degrees

Then;

b^2 = a^2 + c^2 - 2acCosB

8.5^2 = 6.5^2 + 4.5^2 -2(6.5 * 4.5)CosB

72.25 = 62.5 - 58.5CosB

72.5 - 62.5 =  - 58.5CosB

Cos B = -0.1709

B = 100 degrees

Then;

A = 180 - (100 + 31)

A = 49 degrees

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The p-value is less than 0.0001, which is smaller than our level of significance (α = 0.05). Therefore, we reject the null hypothesis and conclude that the mean area of homes built in is not equal to the mean area of homes built in at a 95% confidence level

To determine whether we can conclude that the mean area of homes built in is not equal to that of homes built in, we need to perform a hypothesis test.

Let's set up our hypotheses:

Null hypothesis (H0): The mean area of homes built in is equal to the mean area of homes built in

Alternative hypothesis (Ha): The mean area of homes built in is not equal to the mean area of homes built in

We will use a two-tailed t-test, as we are testing for inequality rather than a specific direction.

Assuming a level of significance (α) of 0.05, our critical values are ±1.96.

To calculate the test statistic, we can use the formula:

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

t = ( - ) / ( / sqrt())

t = -4.72

Using a t-distribution table with degrees of freedom equal to (sample size - 1), we can find the p-value associated with this t-statistic.
The p-value is less than 0.0001, which is smaller than our level of significance (α = 0.05). Therefore, we reject the null hypothesis and conclude that the mean area of homes built in is not equal to the mean area of homes built in at a 95% confidence level. In other words, there is sufficient evidence to suggest that the mean area of homes built in is different from the mean area of homes built in.

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The value of length of AC is,

⇒ AC = 8.25

We have to given that;

CD = 6.6 cm, DE = 3.4 cm, CE = 4.2 cm, and BC = 5.25 cm

Now, We can formulate;

BC / AC = EC / CD

Substitute the values we get;

5.25 / AC = 4.2 / 6.6

Solve for AC;

5.25 x 6.6 / 4.2 = AC

AC = 8.25

Thus, The value of length of AC is,

⇒ AC = 8.25

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Answer: Let X be the random variable denoting the loss due to burglary. Then, we know that X is exponentially distributed with mean 100, which implies that the probability density function of X is given by:

f(x) = (1/100) * exp(-x/100) for x ≥ 0

The probability that a loss is between 40 and 50 can be expressed as:

P(40 ≤ X ≤ 50) = ∫40^50 f(x) dx

= ∫40^50 (1/100) * exp(-x/100) dx

= [-exp(-x/100)]40^50

= exp(-2/5) - exp(-1/2)

Similarly, the probability that a loss is between 60 and r can be expressed as:

P(60 ≤ X ≤ r) = ∫60^r f(x) dx

= ∫60^r (1/100) * exp(-x/100) dx

= [-exp(-x/100)]60^r

= exp(-3/5) - exp(-r/100)

Given that the above two probabilities are equal, we have:

exp(-2/5) - exp(-1/2) = exp(-3/5) - exp(-r/100)

Solving for r, we get:

r = -100 * ln [exp(-2/5) - exp(-1/2) + exp(-3/5)]

r ≈ 85.863

Therefore, the value of r is approximately 85.863.

Using a numerical solver, we can find the value of r: r ≈104.59. So, the value of r is approximately 67.35. To solve this problem, we can use the fact that the exponential distribution is memoryless, which means that the probability of a loss being between two values is the same no matter how much time has elapsed since the last loss.

Let X be the amount of the loss due to burglary. We know that X is exponentially distributed with mean 100, which implies that its probability density function (pdf) is given by f(x) = (1/100) * e^(-x/100) for x > 0.

The probability that a loss is between 40 and 50 is given by the integral of f(x) over that interval:

P(40 < X < 50) = integral from 40 to 50 of f(x) dx
               = integral from 40 to 50 of (1/100) * e^(-x/100) dx
               = e^(-40/100) - e^(-50/100)
               = 0.3679 - 0.3679 * e^(-1/10)
               = 0.0511 (rounded to four decimal places)

Now, let's find r such that the probability that a loss is between 60 and r is also 0.0511. We can set up the following equation:

P(60 < X < r) = integral from 60 to r of f(x) dx
              = e^(-60/100) - e^(-r/100)
              = 0.0511

Solving for r, we get:

e^(-r/100) = e^(-60/100) - 0.0511
r = -100 * ln(e^(-60/100) - 0.0511)
 = 104.59 (rounded to two decimal places)

Therefore, r is approximately 104.59.

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The degrees of freedom and the critical t value (t*) for the given sample size and confidence level of the following are:
(a) df= 5, t*= 1.943; (b) df= 20, t*= 2.845; (c) df= 28, t*= 2.045; (d) df= 11, t*= 3.106.

The critical t value (t*) can be found using the t-distribution table, which takes into account the sample size and confidence level. The degrees of freedom (df) can be calculated using the formula df = n - 1.

(a) n = 6, CL = 90%
df= 5 t*= 1.943

Using the t-distribution table with 5 degrees of freedom and a confidence level of 90%, we find the critical t value to be 1.943.

(b) n = 21, CL = 98%
df= 20 t*= 2.845

Using the t-distribution table with 20 degrees of freedom and a confidence level of 98%, we find the critical t value to be 2.845.

(c) n = 29, CL = 95%
df= 28 t*= 2.045

Using the t-distribution table with 28 degrees of freedom and a confidence level of 95%, we find the critical t value to be 2.045.

(d) n = 12, CL = 99%
df= 11 t*= 3.106

Using the t-distribution table with 11 degrees of freedom and a confidence level of 99%, we find the critical t value to be 3.106.

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The linear regression equation is y = 4,035.33 + 2,115x.

The correlation coefficient is 0.976321113.

The type of correlation is a positive linear correlation.

Yes, the correlation is strong because the correlation coefficient approximately equals to 1.

The amount of Krabby Patties made by Spongebob after working 10 years is $25,185.

In this scenario, the years worked (x) would be plotted on the x-axis of the scatter plot while the patties made (y) would be plotted on the x-axis of the scatter plot.

By critically observing the scatter plot (see attachment) which models the relationship between the years worked (x) and the Patties made (y), an equation for the linear regression is given by:

y = 41,461.54 + 2,714.46x

Next, we would predict the amount of Krabby Patties made by Spongebob after working 10 years as follows;

y = 4,035.33 + 2,115(10)

y = 4,035.33 + 2,1150

y = $25,185.33 ≈ $25,185.

In conclusion, there is a strong correlation between the data because the correlation coefficient (r) approximately equals to 1;

0.7＜|r| ≤ 1   (strong correlation)

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The statement that is NOT true in a Chi-Square test is "Small values of the chi squared test statistic would lead to a decision to reject the null hypothesis."

This is because if the test statistic is small, it means that the observed values are close to the expected values, and there is no significant difference between the populations. Therefore, a small test statistic would lead to a failure to reject the null hypothesis. In a Chi-Square test, we compare the proportions of specified characteristics in different populations, and we wish to determine whether they are the same or not. If the test statistic is large, it means that the observed values are significantly different from the expected values, and we have evidence to reject the null hypothesis. Finally, the P-value will be small if the test statistic is large, indicating strong evidence against the null hypothesis.

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The probability of this event is 3/13. The event described is choosing a red card with an even number from a deck. To express this event as a set, we will include all even-numbered cards from the red suits (diamonds and hearts). This set is: {2♦, 4♦, 6♦, 8♦, 10♦, 2♥, 4♥, 6♥, 8♥, 10♥}.

Now let's find the probability of this event. There are 52 cards in the deck and 10 cards in the event set. Therefore, the probability is:

P(event) = (number of favorable outcomes) / (total number of outcomes) = 10 / 52 = 5 / 26 ≈ 0.1923

The probability of choosing a red card with an even number from a deck is approximately 0.1923 or 19.23%.

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$138.98 was the list price of the sneakers.

Let's use "x" to represent the trainers' list price. Given that the sales tax rate was 7%,

the tax amount corresponded to 0.07x, or 7% of the list price.

The total cost including tax was $148.61, so we can set up the following equation:

x + 0.07x = 148.61

Simplifying and solving for x, we get:

1.07x = 148.61

x = 138.98

Therefore, the list price of the sneakers was $138.98.

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The diameter of the cone is 40 in. Option D

We can see from the diagram shown, that the shape inside the cone is a triangle.

Thus, using the Pythagorean theorem which states that the square of the longest side which is the hypotenuse is equal to the sum of the squares of the other two sides.

We then have that;

36² = 30² + r²

find the squares

1296 = 900+ r²

collect like terms

r² = 396

Find the square root of the sides

r = 19. 89 in

Then,

Diameter = 2(radius)

Substitute

Diameter = 39. 79in

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The inequality that represents the given condition is "p > 12".

Let p be the amount of cheese produced in kg.

Then, the amount of yogurt produced is (p + 24) kg (since it is 24 kg more than cheese). And the amount of ice cream produced is 3p kg (since it is 3 times as much as cheese).

Now, we need to find the possible values of p that satisfy the condition "the dairy farm produced more kilograms of ice cream than yogurt last week." In other words, we need to compare the amount of ice cream produced (3p) with the amount of yogurt produced (p + 24) and ensure that the amount of ice cream is greater.

So, we can write the following inequality:

3p > p + 24

Simplifying this inequality, we get:

2p > 24

Dividing both sides by 2, we get:

p > 12

Therefore, the possible values of p that satisfy the given condition are all values of p greater than 12.

In summary, the inequality that represents the given condition is "p > 12".

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The length of the hypotenuse is 9.5 kilometers

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

Legs = 8,5 and 4.2

The length of the hypotenuse is calculated as

Hyp^2 = Leg 1^2 + Leg 2^2

substitute the known values in the above equation, so, we have the following representation

Hypotenuse^2 = (8.5)^2 + 4.2^2

Evaluate

Hypotenuse^2 = 89.89

So, we have

Hypotenuse = 9.5

Hence, the hypotenuse = 9.5

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The grocer's selection and weighing of the apples is the experiment, while the specific result of the experiment - an average weight of 165 grams - is the outcome.

An event is a combination of one or more outcomes, such as an average weight between 150 and 165 grams. Finally, the average weight of five apples is the numerical representation of an outcome, which is known as the random variable.
Hi! I'd be happy to help explain these terms in the context of your question:

1. Experiment: Selecting and weighing the apples. This is the activity undertaken to gather data.
2. Outcome: The 165 grams average weight. This is a specific result of the experiment.
3. Event: An average weight between 150 and 165 grams. An event is a combination of one or more outcomes.
4. Random variable: The average weight of five apples. This is a numerical representation of an outcome.

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The coordinate system with x-coordinates between -180 and 180 is likely to be the geographic coordinate system, specifically longitude values. This system represents positions on the Earth's surface using two angles: longitude (x-coordinates) and latitude (y-coordinates).

Longitude values range from -180 degrees (180 degrees West) to 180 degrees (180 degrees East) along the Earth's equator. Latitude values, on the other hand, range from -90 degrees (90 degrees South) to 90 degrees (90 degrees North). The origin (0, 0) of this coordinate system is the intersection of the Prime Meridian (Greenwich Meridian) and the Equator.

The units for these coordinates are typically expressed in decimal degrees (DD). These values can be converted into other units such as degrees-minutes-seconds (DMS) or radians, depending on the application or preference. When working with geographic coordinate systems, it is essential to consider the Earth's curvature and potential distortions when calculating distances or areas.

In summary, if a data set has x-coordinates between -180 and 180, the coordinate system is most likely the geographic coordinate system, representing longitude values. The coordinates are typically expressed in decimal degrees, and their positions relate to the Earth's surface.

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Answer:

99

Step-by-step explanation:

132 x ¾ = 396/4

396 divided by 4

99

It is concluded that TU ≅ UV is by the definition of congruence as shown in the solution part.

As per the given figure, the required proof would be as:

Statements:

1. U is the midpoint of SW.

2. T is the midpoint of SU.

3. V is the midpoint of UW.

4. TU || WV (by midpoint theorem)

5. TV || UW (by midpoint theorem)

6. UT = TV (by midpoint theorem)

7. UV = 2VT (by midpoint theorem)

Reasons:

1. Given.

2. Given.

3. Given.

4. Midpoint theorem.

5. Midpoint theorem.

6. Midpoint theorem.

7. Midpoint theorem.

Therefore, TU ≅ UV by the definition of congruence.

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In this scenario, we have a uniformly charged thin rod extending along the x-axis from the origin (x = 0) to positive infinity (x = +∞).

The term "uniformly charged" means that the charge is distributed evenly throughout the entire length of the rod.

To analyze this situation, we can consider the following steps: 1. Determine the linear charge density (λ) of the rod. Since the rod is uniformly charged, λ remains constant along its entire length. λ is usually given in units of charge per length (e.g., coulombs per meter).

2. To find the electric field at a particular point along or outside the rod, we can break the rod into infinitesimally small segments (dx) and consider the contribution of the electric field (dE) from each of these segments.

3. Calculate the electric field (dE) produced by each segment at the desired point using Coulomb's equations , considering the linear charge density (λ) and distance between the segment and the point.

4. Integrate the electric field contributions (dE) from all segments along the entire length of the rod (from x = 0 to x = +∞) to find the total electric field (E) at the point of interest.

By following these steps, you can analyze the electric field and related properties of a uniformly charged thin rod extending along the x-axis from x = 0 to x = +∞.

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A function showing the value of the account after t years is [tex]f(t) = 690(1.021)^{t}[/tex]

The percentage of growth per year (APY) is 21.22%.

In Mathematics and Financial accounting, continuous compounding interest can be determined or calculated by using this mathematical equation (formula):

[tex]f(t) = P_{0}e^{rt}[/tex]

Where:

Based on the information provided above, we can reasonably infer and logically deduce that the function for the future value after t years is given by;

[tex]f(t) = 690(1 + 0.021)^{t}\\\\f(t) = 690(1.021)^{t}[/tex]

Growth per year (APY) =  [tex]e^{r} =e^{0.021}[/tex]

Growth per year (APY) = 1.02122 - 1 = 0.2122 = 21.22%

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A system of linear equations that can be used to determine the number of DVDs Norma purchased and the number of DVDs Lauretta purchased is:

H. n + l = 14

n = 5/2(l)

In order to write a system of linear equations to describe this situation, we would assign variables to the number of DVDs Norma purchased and the number of DVDs Lauretta purchased, and then translate the word problem into an algebraic equation (linear equations) as follows:

Since Norma purchased two and a half times as many DVDs as Lauretta purchased, a linear equation that models the situation is given by;

n = 2 1/2l

n = (5/2)l = 5l/2

Additionally, they purchased a total of 14 DVDs together;

n + l = 14

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Based on the information given, we know that 14% of the sample of 1000 adults reported meeting the daily serving recommendations for fruits and vegetables set by the CDC. To construct a 90% confidence interval, we need to calculate the margin of error.

We can use the formula:

[tex]Margin of error = z* (sqrt(p*(1-p)/n))[/tex]

Where:

z* is the critical value for a 90% confidence interval, which is 1.645
p is the proportion of the sample that met the CDC recommendations, which is 0.14
n is the sample size, which is 1000

Plugging in the values, we get:

Margin of error = 1.645 * (sqrt(0.14*(1-0.14)/1000))
Margin of error ≈ 0.022

Therefore, the answer is e. 0.022.

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The Chain Rule to find dz/ds and Oz/ot is

[tex]Oz/ot = dz/dt = -2s^2 cos(t) sin(t) = -2s^2 cos(t) * (y/x) = -2s^2 y sin(t) / x[/tex]

Thus, [tex]Oz/ot = -2s^2 y sin(t) / x.[/tex]

The chain rule is a rule in calculus that describes how to find the derivative of a composite function. If we have a function that is made up of two or more functions, the chain rule tells us how to find the derivative of the composite function.

To find dz/ds, we need to apply the Chain Rule, which states that if z is a function of u and u is a function of s, then:

dz/ds = dz/du * du/ds

In this case, we have:

[tex]z = x^2 * y^0 = x^2 = (s cos(t))^2 = s^2 cos^2(t)[/tex]

x = s cos(t)

y = s sin(t)

So, we can rewrite z as:

[tex]z = s^2 cos^2(t)[/tex]

Now, let's find du/ds and dz/du:

du/ds = d/ds (s cos(t)) = cos(t)

[tex]dz/du = d/ds (s^2 cos^2(t)) = 2s cos^2(t)[/tex]

Using the Chain Rule formula, we can now find dz/ds:

[tex]dz/ds = dz/du * du/ds = 2s cos^2(t) * cos(t) = 2s cos^3(t)[/tex]

Therefore,[tex]dz/ds = 2s cos^3(t).[/tex]

To find Oz/ot, we need to find dz/dt using the Chain Rule, and then substitute for z in terms of x and y:

dz/dt = dz/dx * dx/dt + dz/dy * dy/dt

[tex]dz/dx = 2xy^0 = 2x = 2s cos(t)[/tex]

[tex]dz/dy = x^2 * 0y^-1 = 0[/tex]

dx/dt = -s sin(t)

dy/dt = s cos(t)

Substituting these values in the Chain Rule formula, we get:

dz/dt = 2s cos(t) * (-s sin(t)) + 0 * (s cos(t)) = -2s^2 cos(t) sin(t)

Now, we can substitute for z in terms of x and y:

[tex]z = x^2 * y^0 = x^2 = (s cos(t))^2 = s^2 cos^2(t)[/tex]

Therefore, we have:

[tex]Oz/ot = dz/dt = -2s^2 cos(t) sin(t) = -2s^2 cos(t) * (y/x) = -2s^2 y sin(t) / x[/tex]

Thus, Oz/ot = -2s^2 y sin(t) / x.

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