12 1 point Suppose P(A) = 0.8, P(B) = 0.5 and P(AUB) = 0.9. Which one of the following statements is true? Events A and B are independent. - Events A and B are both mutually exclusive and independent. The probability of the intersection of A and B is 0.1. Events A and B are mutually exclusive.

The equivalent system of first-order differential equations for the given problem is: 1. dv1/dt = v2 2. dv2/dt = v3 3. dv3/dt = 2v2 - v1 + 1 + t*e ^t

Given differential equation: x''' - 2x'' + x' = 1 + t*e ^t

Step 1: Define new variables.
Let's introduce new variables:
v1 = x'
v2 = v1'
v3 = v2'

Now we have:
v1 = x'
v2 = v1'
v3 = v2'

Step 2: Rewrite the given equation using new variables.
Substitute the new variables into the given differential equation:
v3 - 2v2 + v1 = 1 + t*e ^t

Step 3: Write the equivalent system of first-order differential equations.
Now we have the following equivalent system of first-order differential equations:
dv1/dt = v2
dv2/dt = v3
dv3/dt = 2v2 - v1 + 1 + t*e ^t

So, the equivalent system of first-order differential equations for the given problem is:

1. dv1/dt = v2
2. dv2/dt = v3
3. dv3/dt = 2v2 - v1 + 1 + t*e ^t

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The sinking fund payment will be $1,096.28.
Total payments into the bond will be $61,391.68.
The bond will earn $25,608.32 interest after 14 years.

Let's use the Time Value of Money (TVM) Solver to determine the quarterly payment needed to achieve your goal.

Given:
- Future Value (FV) = $87,000
- Time (n) = 14 years, compounded quarterly, so n = 14 * 4 = 56 quarters
- Interest rate (i%) = 3.2% compounded quarterly, so i% = 3.2 / 4 = 0.8% per quarter
- Present Value (PV) = 0, since we're starting from scratch
- PMT Type: END (payments made at the end of each quarter)

Now, input these values into the TVM Solver:

n = 56
i% = 0.8
PV = 0
PMT = ?
FV = 87,000

Solve for PMT:

PMT = -1,096.28 (rounded to the nearest cent)

The sinking fund payment will be $1,096.28.

To find the total payments into the bond, multiply the payment amount by the number of quarters:

Total payments = PMT * n = 1,096.28 * 56 = $61,391.68

To find the interest earned after 14 years, subtract the total payments from the future value of the bond:

Interest earned = FV - Total payments = 87,000 - 61,391.68 = $25,608.32

So, the bond will earn $25,608.32 in interest after 14 years.

Your answer:
The sinking fund payment will be $1,096.28.
Total payments into the bond will be $61,391.68.
The bond will earn $25,608.32 interest after 14 years.

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Since the line passes through the points (3,-2) and (3, 4), the slope of the line is: C. the slope is undefined.

In Mathematics and Geometry, the slope of any straight line can be determined by using the following mathematical equation;

Slope (m) = (Change in y-axis, Δy)/(Change in x-axis, Δx)

Slope (m) = rise/run

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

By substituting the given data points into the formula for the slope of a line, we have the following;

Slope (m) = (4 + 2)/(3 - 3)

Slope (m) = (6)/(0)

Slope (m) = undefined.

Based on the graph, the slope is the change in y-axis with respect to the x-axis and it is undefined.

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(a) Let X be the length of a cylindrical metal bar. Then, X ~ N(11, 0.25^2), meaning X is normally distributed with mean 11 cm and standard deviation 0.25 cm. To find P(X > 10.5), the probability that a randomly selected cylindrical metal bar has a length longer than 10.5 cm.

To solve this, we can standardize X using the z-score formula:

z = (X - μ) / σ

where μ = 11 (mean) and σ = 0.25 (standard deviation).

So, we have:

z = (10.5 - 11) / 0.25 = -2

Now, we can find the probability using a standard normal distribution table or calculator:

P(X > 10.5) = P(Z > -2) ≈ 0.9772

(b) To find this value, we can use a standard normal distribution table or calculator. First, we need to find the z-score corresponding to the 86th percentile:

P(Z > z) = 0.14

P(Z < z) = 1 - 0.14 = 0.86

Using a standard normal distribution table or calculator, we find that z ≈ 1.08.

Now, we can use the z-score formula to find K:

z = (K - μ) / σ

1.08 = (K - 11) / 0.25

K - 11 = 1.08 * 0.25

K ≈ 11.27

Therefore, the value of K such that there is a 14% chance that a randomly selected cylindrical metal bar has a length longer than K is approximately 11.27 cm.

(c) Mean = $100 + ($80/cm x 11 cm) = $980
Median = $100 + ($80/cm x 11 cm) = $980
Standard deviation = $80/cm x 0.25 cm = $20
Variance = ($80/cm x 0.25 cm)^2 = $400
86th percentile = mean + (1.08 x standard deviation) = $980 + ($20 x 1.08) = $1002.40

(d) The production cost of a cylindrical metal bar has a mean of $980 and a standard deviation of $20. The cost has a variance of $400 and the 86th percentile of the cost distribution is $1002.40.

(e) (II) a lower level than 0.25 cm.

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Using  both and levels of significance and the critical value, we can conclude that the percentage of spam emails sent by the huge firm is different from the percentage in a recent month.

The population proportion of spam emails in a recent month is p = 0.718.

A random sample of emails from a large corporation has a sample proportion of spam emails, p'= 0.645.

We want to test the hypothesis that the population proportion of spam emails in the large corporation is different from p = 0.718.

We will use both 0.05 and 0.01 levels of significance and the critical value method.

To test this hypothesis using the critical value method, we can follow these steps:

The null hypothesis is that the population proportion of spam emails in the large corporation is equal to 0.718:

H0: p = 0.718

The alternative hypothesis is that the population proportion of spam emails in the large corporation is different from 0.718:

Ha: p ≠ 0.718

We will use both 0.05 and 0.01 levels of significance. Since we have a large sample (np > 10 and n(1-p) > 10), we can use the z-test for proportions. The test statistic is calculated as:

z = ( p' - p) / sqrt(p(1-p)/n)

where n is the sample size.

Using a standard normal distribution table, the critical values for a two-tailed test at the 0.05 and 0.01 levels of significance are:

At the 0.05 level: ±1.96

At the 0.01 level: ±2.58

Step 4: Calculate the test statistic and p-value.

Using the formula for the test statistic and the given values, we get:

z = (0.645 - 0.718) / sqrt(0.718(1-0.718)/n)

Since we don't know the population standard deviation, we use the standard error estimated from the sample:

z = (0.645 - 0.718) / sqrt(0.718(1-0.718)/n) = -2.546 / sqrt(0.718(1-0.718)/n)

Using n = 1000 (a reasonable sample size for an email server), we get:

z = -2.546 / sqrt(0.718(1-0.718)/1000) = -9.386

The corresponding p-value for this test statistic is very small (less than 0.0001), indicating strong evidence against the null hypothesis.

At the 0.05 level of significance, the critical value is ±1.96, which does not include the calculated test statistic of -9.386. Therefore, we reject the null hypothesis and conclude that the population proportion of spam emails in the large corporation is different from 0.718.

At the 0.01 level of significance, the critical value is ±2.58, which also does not include the calculated test statistic of -9.386. Therefore, we reject the null hypothesis at this level of significance as well.

In conclusion, we have strong evidence to suggest that the proportion of spam emails in the large corporation is different from the proportion in a recent month (0.718).

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15bags of fiber stuffing will be needed to fill the cones

The volume of one replica ice cream cone is calculated as:

Volume = (1/3) × π × (radius)^2 × height

Radius = 4/2 = 2 inches

Height = 10 inches

Therefore, Volume = (1/3) × π × 2^2 × 10 = 83.78 cubic inches (rounded to two decimal places)

To fill 125 replica cones, the total volume of stuffing required will be:

Total Volume = 125 × 83.78 = 10,472.5 cubic inches

Since each bag of stuffing can fill 720 cubic inches, the number of bags required will be:

Number of bags = 10,472.5 / 720 = 14.55 (rounded up to the nearest whole number)

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The formula for the height above the ground, h(t) is h(t) = 255 sin (π/20 t) + 20.

The amplitude is half the distance between the highest and lowest points, which is (530 - 20)/2 = 255 feet. So A = 255.

The period is 40 minutes, so B = 2π/40 = π/20.

At t = 0 (when we board the Ferris Wheel), we are 20 feet above the ground.

This means there is no phase shift, so C = 0.

The vertical shift is also 20 feet, so D = 20.

Putting it all together, we have:

h(t) = 255 sin (π/20 t) + 20

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In the given triangle, the measure of angle A is approximately 55°

From the question, we are to determine the measure of angle A

To determine the measure of angle A, we will use SOH CAH TOA

sin (angle) = Opposite / Hypotenuse

cos (angle) = Adjacent / Hypotenuse

tan (angle) = Opposite / Adjacent

Thus,

We can write that

sin (A) = BC / AB

First, we will determine the length of BC

From the Pythagorean theorem,

BC² = AB² - AC²

BC² = 37² - 21²

BC² = 928

BC = √928

BC = 4√58

Thus,

sin (A) = (4√58) / 37

sin (A) = 0.8233

A = sin⁻¹ (0.8233)

A = 55.4165°

A ≈ 55°

Hence,

The measure of angle A is 55°

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If the product of the terms of the fraction is 162 and the fraction is equivalent to 2/9 then the fraction is 6/27.

Let's assume the two terms of the fraction are x and y. We are given that xy = 162. Also, we know that the fraction is equivalent to 2/9. So, we can write,

x/y = 2/9

Cross-multiplying, we get,

9x = 2y

Solving for y, we get,

y = (9/2)x

Substituting this value of y in the equation xy = 162, we get,

x(9/2)x = 162

Simplifying, we get,

(9/2)x² = 162

x² = 36

x = 6 (taking the positive square root as x and y are positive)

Substituting this value of x in the equation y = (9/2)x, we get,

y = (9/2)(6) = 27

So, the fraction is 6/27, which simplifies to 2/9.

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Complete question - The product of two terms of a fraction is 162, find the fraction if it is equivalent to 2/9.

Scatterplot would  more appropriate for displaying and describing the relationship between the age and the number of vocabulary words.

We have to compare scatterplot and line graph.

The association between age and vocabulary size would be better represented and explained by a scatter plot.

The link between two continuous variables is shown using a scatter plot; in this instance, age is a continuous variable whereas the quantity of vocabulary items is a discrete variable.

The relationship between the two variables can be visually examined with the use of scatter plots, which also reveal the direction and strength of the association.

Thus, the scatterplot is best choice.

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To determine the crucial values and create the confidence intervals for the mean, we may utilise the t-distribution and t-score.

(a) The variable of interest here is the average number of jobs that retired men had during their lifetimes.

(b) To find the 92% confidence interval of the mean number of jobs, we can use the formula:

CI = X ± Z * (σ / √n)

where X is the sample mean, Z is the Z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.

Using the given values, we have:

X = 6.6

Z = Z-score corresponding to 92% confidence level (which can be found using a standard normal distribution table or calculator)

σ = 2.1

n = 54

(c) To find the 96% confidence interval of the mean number of jobs, we can use the formula:

Confidence Interval = Sample Mean ± Margin of Error

The margin of error can be calculated using the formula:

Margin of Error = Critical Value * Standard Error

First, we need to determine the critical value corresponding to a 96% confidence level. Since the sample size is relatively large (n > 30), we can use the Z-distribution. The critical value can be found by looking up the z-score corresponding to a confidence level of 96% in the standard normal distribution table or using a statistical calculator. For a 96% confidence level, the critical value is approximately 1.750.

Next, we need to calculate the standard error of the mean. The standard error can be computed using the formula:

Standard Error = Population Standard Deviation / √(Sample Size)

Given that the population standard deviation is 2.1 and the sample size is 54, we can plug these values into the formula:

Standard Error = 2.1 / √(54)

Calculating this, we find that the standard error is approximately 0.285.

Now we can calculate the margin of error:

Margin of Error = 1.750 * 0.285

The margin of error is approximately 0.499.

Finally, we can construct the confidence interval:

Confidence Interval = Sample Mean ± Margin of Error

Confidence Interval = 6.6 ± 0.499

Therefore, the 96% confidence interval of the mean number of jobs is approximately (6.101, 7.099).

(d) The 96% confidence interval will be smaller than the 92% confidence interval.

This is because as the confidence level increases, the range of the confidence interval becomes wider. A higher confidence level requires a larger interval to capture a greater proportion of the population. Therefore, the 96% confidence interval will be wider than the 92% confidence interval, indicating a larger range of plausible values for the population mean.

(e) To compute the confidence intervals, we use the t-test method. The assumptions we need to make are:

Random Sampling: The sample should be a simple random sample from the population.

Normality: The population should follow a normal distribution, or for larger sample sizes (typically n > 30), the sampling distribution of the sample mean should be approximately normal due to the central limit theorem.

Independence: The observations in the sample should be independent of each other.

Homogeneity of Variance (Optional): If comparing two or more groups, the population variances should be equal. This assumption is not necessary when constructing a confidence interval for a single population mean.

Under these assumptions, we can use the t-distribution and the t-score to calculate the critical values and construct the confidence intervals for the mean.

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The correct range of the given function is :

c. {y € R, y>=0}.

To determine the range of the function y = (x-2)^(1/2), we need to consider the possible values of y that can be obtained for different values of x.

a. {x € R} means that x can take any real value. However, since the square root of a negative number is not a real number, y can only take non-negative values. So, the range is {y € R, y>=0}.

b. {x € R, x>=2} means that x can take any real value greater than or equal to 2. Again, the square root of a negative number is not a real number, so y can only take non-negative values. So, the range is {y € R, y>=0}.

d. {y € R} means that y can take any real value. However, if we plug in a value of x less than 2, we get a negative value under the square root, which is not a real number. So, the range is not {y € R}.

Therefore, the correct answer is c. {y € R, y>=0}.

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The limit of the ratio of the probabilities is[tex]$\frac{3}{4}$[/tex].

Given: [tex]$\$ P(Z=0)=p \$[/tex], [tex]\$ P(Z=1)=q \$[/tex], [tex]\$ P(Z=3)=r \$$[/tex], where[tex]$\$ p+q+r=1 \$$[/tex] and [tex]$\$ E[Z] < 1 \$$[/tex].

(a) To find the recursion formula for [tex]$\$ W_{-} a(u) \$$[/tex], we use the following formula:

[tex]$$W_a(u)=P(Z=a)+\sum_{k=0}^{u-1} P(Z=a+3 k) W_a(u-1-k)$$[/tex]

where [tex]$\$ \mathrm{a} \$$[/tex] is a non-negative integer and [tex]$\$ \mathrm{u} \$$[/tex] is a positive integer.

Using the given probabilities, we have:

[tex]$$\begin{aligned}& W_0(u)=p+\sum_{k=0}^{u-1} r W_0(u-1-k) \\& W_1(u)=q+\sum_{k=0}^{u-1} p W_1(u-1-k)+\sum_{k=0}^{u-1} r W_1(u-1-k-1) \\& W_3(u)=r+\sum_{k=0}^{u-1} q W_3(u-1-k)\end{aligned}$$[/tex]

(b) To find [tex]$\$ 19$[/tex], we use the fact that [tex]$\$ W_{-} O(u)+W_{-} 1(u)+W_{-} 3(u)=1 \$$[/tex] for all positive integers [tex]$\$[/tex] u [tex]\$$[/tex].

We take the limit as [tex]$\$$[/tex] u [tex]$\$$[/tex] approaches infinity:

[tex]$$\lim _{u \rightarrow \infty} W_0(u)+\lim _{u \rightarrow \infty} W_1(u)+\lim _{u \rightarrow \infty} W_3(u)=1$$[/tex]

Since [tex]$\$ E[Z] < 1 \$$[/tex], we have. Also, from part (a), we have:

[tex]$$\begin{aligned}& \lim _{u \rightarrow \infty} W_0(u)=\lim _{u \rightarrow \infty} W_0(u-1) \\& \lim _{u \rightarrow \infty} W_1(u)=p \lim _{u \rightarrow \infty} W_1(u-1)+\lim _{u \rightarrow \infty} W_1(u-2)\end{aligned}$$[/tex]

solve for the limits as:

[tex]$$\begin{aligned}& \lim _{u \rightarrow \infty} W_0(u)=\frac{p}{1-r} \\& \lim _{u \rightarrow \infty} W_1(u)=\frac{q p}{1-p-r}\end{aligned}$$[/tex]

Therefore, we have:

[tex]$$\lim _{u \rightarrow \infty} f(u)=\lim _{u \rightarrow \infty} \frac{W_1(u)}{W_0(u)+W_1(u)}=\frac{q p}{p+(1-r)}=\frac{3}{4}$$[/tex]

Thus, the limit of the ratio of the probabilities is[tex]$\frac{3}{4}$[/tex].

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The equation without decimal and fraction is 20x = -39.

Given is an equation 5x/6 +2 = 3/8

So, [tex]\frac{5x}{6} +2 = \frac{3}{8} \\\\[/tex]

Multiply the equation by 48,

[tex]\frac{5x}{6} +2 = \frac{3}{8} \\\\40x + 96 = 18\\\\40x = -78\\\\\\20x = -39[/tex]

Hence the equation without decimal and fraction is 20x = -39.

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Answer:

The solutions to the quadratic equation y = x^2 + 3x + 4 are (-1.5 + 1.936i) and (-1.5 - 1.936i).

Step-by-step explanation:

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The table showing the exponential function is completed and presented below  

x      y

0    96

1    192

2   384

3   768

The expression for the table is given as

y = a * b^x,

Point (0, 96) is on the table.

a = 96,

when b = 2, we have that

for x = 1

y = 96 x 2^x = 96 x 2^1 = 192

for x = 2

y = 96 x 2^x = 96 x 2^2 = 384

for x = 3

y = 96 x 2^x = 96 x 2^3 = 768

Therefore, The required exponential function will be  

x      y

0    96

1    192

2   384

3   768

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The difference of three times a number x and six is greater than or equal to the sum of fifteen and twenty-four times the number is 3x-6≥15+24x

The difference of three times a number and six is greater than or equal to the sum of fifteen and twenty four times a number

Difference is subtraction and sum is nothing but addition

Let the number be x.

The given sentence is changed to the expression or inequality as given below.

3x-6≥15+24x

Hence, the difference of three times a number x and six is greater than or equal to the sum of fifteen and twenty-four times the number is 3x-6≥15+24x

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The frequency of this wave is 1 s⁻¹.

Since, We know that;

Frequency describes the number of waves that pass a fixed place in a given amount of time.

Given that;

A coiled spring with coils that are closely spaced then widely then closely then widely then closely, ending with a yellow line labeled 1 second.

We know that -

Frequency = 1 / Time period

f = 1/T

f = 1/1

f = 1 s⁻¹

Therefore, the frequency of this wave is 1 s⁻¹.

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Since there is no way to list all the infinite sequences of 0s and 1s, we have proved that S is uncountable.

To show that S is uncountable, we need to show that there is no one-to-one correspondence between the set of infinite sequences of 0s and 1s and the set of natural numbers (which are used to count elements in a set).
We can use the diagonal argument, which is similar to the one used to show that the real numbers are uncountable.
Assume that S is countable, meaning that there is a way to list all the infinite sequences of 0s and 1s as a sequence: s1, s2, s3, ...
We can construct a new sequence t by choosing the opposite of each digit on the diagonal of the previous sequences: if the diagonal digit of s1 is 0, then the first digit of t is 1, and vice versa. Similarly, if the diagonal digit of s2 is 1, then the second digit of t is 0, and vice versa. And so on for all the diagonal digits.
The sequence t is guaranteed to be different from all the sequences in the original list, because it differs from each sequence at least at one position (the diagonal position). Therefore, t cannot be in the original list, which means that the original list did not include all the infinite sequences of 0s and 1s.
Since we have shown that there is no way to list all the infinite sequences of 0s and 1s, we have proved that S is uncountable.

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The different ways to choose the ones digit A are:

- A can be any digit if the sum of the first 6 digits is divisible by 3.
- A can be 2, 5, or 8 if the sum of the first 6 digits leaves a remainder of 1 when divided by 3.
- A can be 1, 4, 7, or 0 if the sum of the first 6 digits leaves a remainder of 2 when divided by 3.

To determine if a number is divisible by 3, we can add up the digits and check if the sum is divisible by 3. For the number 572435A, the sum of the first 6 digits is 5 + 7 + 2 + 4 + 3 + 5 = 26. We need to add a number A to this sum to make it divisible by 3.

There are three possible scenarios to make the sum divisible by 3:

1. If the sum of the first 6 digits is already divisible by 3, then any digit A can be added to the end to make the number divisible by 3. In this case, the sum of the first 7 digits will also be divisible by 3.

2. If the sum of the first 6 digits leaves a remainder of 1 when divided by 3, then A must be 2, 5, or 8. Adding any other digit to the end will result in a sum that is not divisible by 3. In this case, adding 2, 5, or 8 to the end of the number will result in a sum of digits that is divisible by 3.

3. If the sum of the first 6 digits leaves a remainder of 2 when divided by 3, then A must be 1, 4, 7, or 0. Adding any other digit to the end will result in a sum that is not divisible by 3. In this case, adding 1, 4, 7, or 0 to the end of the number will result in a sum of digits that is divisible by 3.

Therefore, the different ways to choose the ones digit A in the number 572435A so that the number will be divisible by 3 are:

- A can be any digit if the sum of the first 6 digits is divisible by 3.
- A can be 2, 5, or 8 if the sum of the first 6 digits leaves a remainder of 1 when divided by 3.
- A can be 1, 4, 7, or 0 if the sum of the first 6 digits leaves a remainder of 2 when divided by 3.

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the dimensions of the rectangle are L = 15 inches and W = 15 inches, and the minimum perimeter is:    P = 2L + 2W = 60 inches.

Let the length and width of the rectangle be L and W, respectively, so that the area of the rectangle is A = LW = 225. We want to find the dimensions of the rectangle with minimum perimeter P = 2L + 2W, and then find the minimum perimeter.

Using the given area, we can solve for one of the variables in terms of the other:

L = 225/W

Substituting this expression for L into the expression for the perimeter, we get:

P = 2(225/W) + 2W

Taking the derivative of P with respect to W and setting it equal to zero to find the minimum, we get:

[tex]dP/dW = -450/W^2 + 2 = 0[/tex]

Solving for W, we get:

W^2 = 225

Since W must be positive (it is a length), we take the positive square root:

W = 15

Substituting this value of W back into the expression for L, we get:

L = 225/W = 15

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The question is solved using two linear equations. By analyzing the money value and number of coins, it is determined that Adriel has 2 nickels and 5 dimes.

In this problem, you need to understand that a nickel is worth 0.05 dollars (or 5 cents) and a dime is worth 0.10 dollars (or 10 cents). Adriel has a total of $0.60. Define x as the nickels and y as the dimes. You can set up the following two equations based on the information:

If we multiply the second equation by 0.05, we get 0.05x + 0.05y = 0.35. Subtract this from the first equation, which gives 0.05y = 0.25 and solving this gives y = 5. Substituting y = 5 into x + y = 7, we get x = 2. Therefore, Adriel has 2 nickels and 5 dimes.

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The 95% confidence interval for the population mean is approximately (61.91 inches, 64.89 inches)

To construct the 95% confidence interval for the population mean, we will use the given information and the formula:

[tex]Lower Bound = Point Estimate - (1.96)(\frac{s}{\sqrt{n} } )[/tex]
[tex]Lower Bound = Point Estimate +(1.96)(\frac{s}{\sqrt{n} } )[/tex]

In this case, the point estimate is the mean height of the sample, which is 63.4 inches. The standard deviation (s) is 2.4, and the sample size (n) is 10. Now we can plug these values into the formula:

[tex]Lower Bound = 63.4 - (1.96)\frac{2.4}{\sqrt{10} }  = 63.4 - (1.96)(0.759) = 63.4 - 1.489 = 61.91[/tex]
[tex]Upper Bound = 63.4 + (1.96)\frac{2.4}{\sqrt{10} }  = 63.4 + (1.96)(0.759) = 63.4 + 1.489 = 64.89[/tex]

Therefore, the 95% confidence interval for the population mean is approximately (61.91 inches, 64.89 inches).

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Answer: Whole thing=20.5

First half (17,18,19,20)=18.5

Second half (21,24,25,27) = 24.5

Step-by-step explanation:

eliminate numbers on both sides till you get to the middle if there is an even number add up the two numbers in the middle and divide them by 2

For example, In the whole thing, you are left with 20 and 21 so

20+21 =41/2= 20.5

For example, In the first part, you are left with 18 and 19 so

18+19 =37/2= 18.5

For example, In the first part, you are left with 24 and 25 so

24+25 =49/2= 24.5

I think it's 9, tell me if I'm wrong

The equation to show the number of cakes that can be baked with 50 pounds of flour, is 181. 5 = c × 112.

If we represent the quantity of cakes Betty can make as "c", and it is known that each cake requires 112 cups of flour, with a total of 181.5 cups available, then the equation may be expressed as:

Total flour = Number of cakes × Flour per cake

181. 5 = c × 112

c = 181. 5 / 112

c = 1.62 cakes

In conclusion, 1. 62 cakes can be baked.

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The given procedure does result in a binomial distribution. This is because it meets the necessary conditions for a binomial distribution:

1. Fixed number of trials: There are 74 trials (rolling the die 74 times).
2. Two outcomes: The outcome of each roll is either odd (success) or even (failure).
3. Independent trials: The outcome of one roll does not affect the outcome of any other roll.
4. Constant probability: The probability of rolling an odd number remains the same for each roll (1/2, since there are 3 odd numbers out of 6 sides).'

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The balance after 3 years is $552.50.

We have,

To find the interest earned, we can use the simple interest formula:

I = Prt

where I is the interest earned, P is the principal (starting amount), r is the interest rate (as a decimal), and t is the time (in years).

Substituting the given values, we get:

I = 500 x 0.035 x 3

I = $52.50

Therefore, the interest earned is $52.50.

To find the balance, we need to add the interest earned to the principal:

Balance = Principal + Interest

Balance = $500 + $52.50

Balance = $552.50

Therefore,

The balance after 3 years is $552.50.

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(24 ÷ 3) + x expresses the expression that represents the quotient of 24 and 3 plus x.

The expression refers to a mathematical phrase with two or more numbers or variables with mathematical operations such as addition, subtraction, division, multiplication, exponential, and so on. Examples of expression include 2a + 3p, and 9p.

To convert the given phrase into the expression, we have to start with the first operation which is a division that is represented by the word quotient. Thus we get 24 ÷ 3

Then we add the operation of addition which is represented by the word plus to the existing equation and we get (24 ÷ 3) + x

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Answer:

Step-by-step explanation: