1)A factory makes propeller drive shafts for ships. A quality assurance engineer at the factory needs to estimate the true mean length of the shafts. She randomly selects four drive shafts made at the factory, measures their lengths, and finds their sample mean to be 1000 mm. The lengths are known to follow a normal distribution whose standard deviation is 2 mm. Calculate a 95% confidence interval for the true mean length of the shafts. Input your answers for the margin of error, lower bound, and upper bound.
a)Determine Margin of Error for this 95% confidence interval.
b)Input the lower bound. (Round to three decimal places)
c)Input the upper bound. (Round to three decimal places)
2)To assess the accuracy of a laboratory scale, a standard weight that is known to weigh 1 gram is repeatedly weighed a total of n times and the mean of the weighings is computed. Suppose the scale readings are normally distributed with unknown mean μ and standard deviation σ = 0.01 g. How large should n be so that a 95% confidence interval for μ has a margin of error of ± 0.0001?
3)A medical researcher is working on a new treatment for a certain type of cancer. The average survival time after diagnosis for the standard treatment is two years. So the null hypothesis is that average survival time after diagnosis is the same for the new treatment and the standard treatment.
In an early trial, she tries the new treatment on three subjects, who have an average survival time after diagnosis of 4.5 years. Even though the sample is small, the results are statistically significant at the 0.05 significance level. Consequently, she rejects the null hypothesis.
In a future study, it is determined that the new treatment does not increase the mean survival time in the population of all patients with this particular type of cancer. The researcher has
Committed a type I error.
Incorrectly used a 0.05 significance test when she should have computed the P-value.
Incorrectly used a 0.05 significance level when she should have used a 0.01 significance level.
Committed a type II error.
4)If the level of significance, α{"version":"1.1","math":"\alpha"}, is made very small, thereby making the probability of committing a Type 1 error very small, what happens to the probability of committing a Type 2 error?
By reducing the probability of committing a Type 1 error, we increase the probability of committing a Type 2 error.
There is no specific relationship between the two probabilities.
By reducing the probability of committing a Type 1 error, we also reduce the probability of committing a Type 2 error.
The relationship between the two probabilities depends on how the study is set up.

If the null hypothesis were correct, there would be a 0.47 percent chance of getting a sample proportion that deviates from the population proportion's hypothesised value

Proportion refers to the relationship between two quantities or numbers, indicating how they are related to each other in size or amount.

The best interpretation of a Z score of 0.47 for a sample proportion is:

b. If the null hypothesis were correct, there would be a 0.47 percent chance of getting a sample proportion that deviates from the population proportion's hypothesised value.

This interpretation is in line with the definition of a Z score, which is a measure of how many standard deviations a sample statistic (in this case, the sample proportion) is not what would be anticipated if the null hypothesis were true. A Z score of 0.47 means that the sample proportion is 0.47 standard deviations away from the expected value under the null hypothesis. Therefore, the interpretation that the probability of obtaining a sample proportion as far or farther than observed from the hypothesized value of the population proportion is 0.47, assuming the null hypothesis is true, is the most appropriate.

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As a result, Lara made a $6,262.71 initial deposit into her savings account.

A credit card user who pays 12% interest (or any other loan type that charges compound interest) will double their debt in six years. The rule can also be applied to determine how long it takes for inflation to cause money's value to decrease by half.

To calculate the initial investment, we can apply the continuous compounding formula:

A = Pe(rt)

Where:

A = the current balance ($7,000.00)

P = the initial deposit (unknown)

r = the annual interest rate (11% or 0.11 as a decimal)

t = the time in years (1 year)

Plugging in these values, we get:

$7,000.00 = Pe(0.11 * 1)

A shorter version of the exponential expression:

$7,000.00 = Pe0.11

$7,000.00 = P * 1.1166 (rounded to 4 decimal places)

Dividing both sides by 1.1166:

P = $6,262.71 (rounded to the nearest cent)

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The fraction of the guests are 10 or 11 years old is 5/6

A fraction represents a part of a whole, where the whole is divided into equal parts.

To find the fraction of guests who are 10 or 11 years old, we need to add the fractions of guests who are 10 years old and 11 years old.

3/12 of the guests are 10 years old, which can be simplified to 1/4.

7/12 of the guests are 11 years old.

Therefore, the fraction of guests who are 10 or 11 years old is

1/4 + 7/12 = 3/12 + 7/12

Add the fractions

= 10/12

Simplify the fraction

= 5/6

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(a) To predict how many supermarkets are expected to adopt the new procedure over a long period of time, we can analyze the behavior of the differential equation using a phase portrait.

The equation can be rewritten as dN/N = (1-0.0005N)dt. Integrating both sides, we get ln|N| = t - 0.0005N^2/2 + C, where C is the constant of integration. Solving for N, we have:

N(t) = +/- sqrt((2ln|N| - 2C)/0.001)

We can see that the solutions are of the form of a hyperbola, with N approaching the asymptotes y=0 and y=2000. The equilibrium point is N=0, which is unstable, and the critical point is N=2000, which is stable.

Therefore, over a long period of time, we expect the number of supermarkets using the computerized checkout system to approach 2000.

(b) To solve the initial-value problem, we can use the separation of variables:

dN/N = (1-0.0005N)dt

ln|N| = t - 0.00025N^2 + C

N(0) = 1

Substituting N=1 and t=0, we get C=0. Therefore, the solution is:

ln|N| = t - 0.00025N^2

N = e^(t-0.00025N^2)

Using a graphing utility, we can plot the solution curve for N(t):

The graph confirms that the solution curve approaches 2000 as t increases.

When t=15, we can evaluate N(15) using the solution:

N(15) = e^(15-0.00025N^2)

Rounding to the nearest integer, we get N(15) = 1719.

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Let m and n be two arbitrary integers. We want to prove that the relation R is an equivalence relation, i.e. it is reflexive, symmetric, and transitive.

Reflexive:  We must show that mRm for all m ∈ Z.

Since the difference of m and m is 0, which is an even number, we have mRm.

Therefore, the relation R is reflexive.

Symmetric: We must show that if mRn, then nRm.

Let mRn, i.e., the difference of m and n is an even number.

Then the difference of n and m is also an even number.

Therefore, nRm, and the relation R is symmetric.

Transitive: We must show that if mRn and nRp, then mRp.

Let mRn and nRp, i.e., the difference of m and n is an even number and the difference of n and p is also an even number.

The sum of the difference of m and n and the difference of n and p is the difference of m and p, which is an even number.

Therefore, mRp, and the relation R is transitive.


Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

Conclusion: The relation R is an equivalence relation.

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a) Sample size if the lower control limit is to be nonzero: 50
b) Sample size if the probability of detecting a shift to 0.04 is to be 0.50: 100

a) How large should the sample size be if the lower control limit is to be nonzero?

n = (2σ / d)²We know that:

Center line (CL) = 0.01

Sigma (σ) = LCL = 0.005

d = Centerline - LCL = 0.01 - 0.005 = 0.005

Substituting the values in the formula, we get

n = (2 * 0.005 / 0.01)²= 50 Hence, if the lower control limit is to be nonzero, the sample size should be 50.

b) How large should the sample size be if we wish the probability of detecting a shift to 0.04 to be 0.50?

The probability of detecting a shift to 0.04 is denoted by β and is calculated using the following formula:

β = Φ [(-Zα/2 + Zβ) / √ (p₀q₀/n)], Where, Φ is the standard normal distribution function, Zα/2 is the critical value for the normal distribution at the (α/2)th percentile, Zβ is the critical value for the normal distribution at the βth percentile, p₀ is the assumed proportion of nonconforming items, q₀ is 1 – p₀, and n is the sample size.

In order to determine the sample size, we must first select a value for β. If we select a value for β of 0.50, then β = 0.50. This implies that we have a 50% chance of detecting a shift if one occurs. Since the exact value for p₀ is unknown, we assume that p₀ = 0.01, which is equal to the center line.

n = (Zα/2 + Zβ)² p₀q₀ / β², Substituting the values in the formula, we get,

n = (Zα/2 + Zβ)² p₀q₀ / β²= (1.96 + 0.674)² (0.01) (0.99) / 0.50²= 99.7 ≈ 100

Hence, if we wish the probability of detecting a shift to 0.04 to be 0.50, the sample size should be 100.

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The slopes of the two line are equal. Hence, the two lines are parallel.

A line's slope is a gauge of the line's steepness. The ratio of the vertical change (change in y) to the horizontal change (change in x) between any two locations on the line is what is meant by this term. When a line moves from left to right, the slope might be positive, negative, zero, or undefined. When a line moves from left to right, the slope can be negative (when the line is vertical). The slope is determined using the following formula and is represented by the letter m:

m = (y2 - y1) / (x2 - x1)

where (x1, y1) and (x2, y2) are any two points on the line.

Given that, the first line passes through the points (-5, -5) and (-3, -3).

The slope is given by:

slope = (change in y) / (change in x)

slope = (-3 - (-5)) / (-3 - (-5)) = 1

The second line passes through the points (3, 1) and (4, 2).

slope = (2 - 1) / (4 - 3) = 1

The slopes of the two line are equal. Hence, the two lines are parallel.

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Answer: Starting with the given equation:

(cos θ)(cos θ) + 1 = (sin θ)(sin θ)

We can use the identity cos² θ + sin² θ = 1 to rewrite the right-hand side:

(cos θ)(cos θ) + 1 = 1 - (cos θ)(cos θ)

Combining like terms, we get:

2(cos θ)(cos θ) = 0

Dividing both sides by 2, we get:

(cos θ)(cos θ) = 0

Taking the square root of both sides, we get:

cos θ = 0

This equation is true for θ = π/2 + kπ, where k is any integer. So the solutions to the equation are:

θ = π/2 + kπ, where k is any integer.

Enjoy!

Step-by-step explanation:

The journal entry in the stokvel's ledger to record the distribution of the Nestle Cremora boxes to Mrs. Bosoga would be as follows:

Date Account Debit Credit

Dec 2022 Nestle Cremora ZAR 3,750

Dec 2022 Mrs. Bosoga  ZAR 3,750

The Nestle Cremora account is debited with ZAR 3,750, representing the cost of the 15 boxes of Nestle Cremora (15 boxes x ZAR 250 per box). The Mrs. Bosoga account is credited with the same amount, indicating that she has received the Nestle Cremora boxes.

The impact on the stokvel's balance sheet would be a decrease in the value of the Nestle Cremora inventory by ZAR 3,750, which would be reflected as a reduction in the stokvel's assets.

The impact on the income statement would be negligible, as the distribution of the Nestle Cremora boxes would not result in any income or expense for the stokvel.

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Mrs. Bosoga is a member of a stokvel, a savings club where members contribute money regularly and receive payouts periodically. During December 2022, Mrs. Bosoga received a share of 15 boxes of Nestle Cremora from the stokvel. The market value of each box of Nestle Cremora at the time was ZAR 250. The stokvel keeps track of its transactions using a ledger. What would be the journal entry in the stokvel's ledger to record the distribution of the Nestle Cremora boxes to Mrs. Bosoga? Also, what would be the impact on the stokvel's balance sheet and income statement?

Answer:

  right angle

Step-by-step explanation:

You want to know the kind of angle formed between a radius and a tangent.

A tangent to a circle is always perpendicular to the radius at the point of tangency.

The angle is a right angle.

Step-by-step explanation:

kaitlyn score is   6 points above the mean

                                 z-score =  6 / 12   = .5

kiersten score is 34.8 above the mean   z-score = 34.8/29 = 1.2

rebecca score is .54 above the mean   z -score = .54/ .3 = 1.8

rebecca scored the highest percentile (highes z-score)  of the three....the best

Given:

81x = 243x

= 243 / 81x

= 3

x = 3

Answer:

4,320 times

Step-by-step explanation:

12 x 6 = 72 beats per minute

72 x 60 = 4320 beats in 60 minutes

Answer: 840 in 30 minuets

Step-by-step explanation:

As per the volume, the dimension of an Extra Large Box is 3.78 feet.

Let's call the length of one side of the cube "s". Since the volume of the cube is given as 512 cubic feet, we can set up an equation to relate the volume to the length of one side:

Volume of cube = s³ = 512 cubic feet

To solve for "s", we can take the cube root of both sides of the equation:

s = ∛512

We can simplify this expression by finding the prime factorization of 512:

512 = 2⁹

Therefore, we can rewrite the expression for "s" as:

s = ∛2⁹

Using the properties of exponents, we know that the cube root of 2^9 is the same as 2 raised to the power of (1/3) times 9:

s = [tex]2^{1/3} \times 9^{1/3}[/tex]

We can simplify this expression further by recognizing that 9 is a perfect cube, and its cube root is 3:

s = [tex]2^{1/3} \times 3[/tex]

Therefore, the length of one side of the cube-shaped box is:

s = [tex]2^{1/3} \times 3[/tex] feet

Since all sides of the cube are equal in length, the dimensions of the box are:

Length = Width = Height = [tex]2^{1/3} \times 3[/tex] feet = 3.78 feet.

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Complete Question:

Head Stevedore loads Extra Large Boxes, also in the shape of perfect cubes. The volume of each box is 512 cubic feet. What are the dimension of an Extra Large Box?

The pair that is the mean and standard error of x is [65, 2.236]. So, the correct option is a.

Suppose a random sample of 80 measurements is selected from a population with a mean of 65 and a variance of 300. We are required to select the pair that is the mean and standard error of x. The standard error of the mean (SEM) is calculated as follows :

$$SEM = \frac{\sigma}{\sqrt{n}}$$

Where σ is the standard deviation, and n is the number of observations or sample size.

Given that variance,

σ2 = 300

Therefore,

σ = √300 = 17.32.

Substituting the values in the formula, we have

$$SEM = \frac{\sigma}{\sqrt{n}} = \frac{17.32}{\sqrt{80}} = 1.9365$$

Therefore, the mean and standard error of x is [65, 1.936]. Option A is not the answer because the value of the standard error in that option is incorrect. Option B is not the answer because the value of the standard error in that option is incorrect. Option C is not the answer because the value of the standard error in that option is incorrect.

Option D is not the answer because the first value in that option is not the mean of x. Option E is incorrect because the value of the standard error is incorrect.

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The solution is, (a/2) ^4  or a^4/16 is in Simplest Form of (8a^3)^-4/3.

An exponent refers to the number of times a number is multiplied by itself. For example, 2 to the 3rd (written like this: 23) means: 2 x 2 x 2 = 8. 23 is not the same as 2 x 3 = 6. Remember that a number raised to the power of 1 is itself.

here , we have,

(8a^-3)^-4/3

split into two parts

8^ -4/3   *  (a^-3)^-4/3

using the power to the power rule we can multiply the exponents

8^(-4/3)  *a^(-3*-4/3)

8^ (-4/3) * a^(4)

replace 8 with 2^3

(2^3)^(-4/3) * a^(4)

using the power to the power rule we can multiply the exponents

2^(3*-4/3) * a^(4)

2 ^ (-4) * a^4

the negative exponent means it goes in the denominator if it is in the numerator

a^4/2^4

make a fraction

(a/2) ^4

or a^2/16

Hence, The solution is, (a/2) ^4  or a^4/16 is in Simplest Form of (8a^3)^-4/3.

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Answer:

0

Step-by-step explanation:

0

Thank me...........

[tex]x=30^o,270^o \ [0^0\leq x\leq 360^0][/tex]

Explanation:

We know,

[tex]cos2x=cos^2 \ x-sin^2 \ x=1-2sin^2 \ x[/tex]

So, let's solve the equation now,

[tex]sin \ x=cos2x=1-2 \ sin^2 \ x[/tex]

[tex]\longrightarrow \ 2 \sin^2 \ x+sin \ x-1=0[/tex]

[tex]\longrightarrow \ 2 \sin^2 \ x+2\sin x-sin \ x-1=0[/tex]

[tex]\longrightarrow \ 2\sin \ x(sin\ x+1)-1(sin \ x+1)=0[/tex]

[tex]\longrightarrow \ (2\sin \ x-1)(sin \ x+1)=0[/tex]

Now,

[tex]2\sin \ x-1=0[/tex]

[tex]\longrightarrow \ sin \ x=\dfrac{1}{2}[/tex]

[tex]\longrightarrow x=sin^{-1}(\dfrac{1}{2})[/tex]

[tex]\longrightarrow x=30^o[/tex]

And, [tex]sin \ x+1=0[/tex]

[tex]\longrightarrow x=sin^{-1}(-1)=270^o[/tex]

As we just need the general solutions, we should take only this two values as the general solutions.

Answer:

[tex]30^0,270^0[/tex]

That's it!

The height of the roller coaster is 199.9 feet.

According to the information provided,

There are two methods you can use to find the height of the roller coaster.

Method 1: Trigonometric ratios

To use this method, we can use the fact that the angle between the hypotenuse and the height of the triangle measures 45 degrees.

We can thus use the trigonometric ratio for the sine of 45 degrees, which is √(2)/2.

Setting up the equation, we have,

⇒ sin(45 degrees) = x/283

Solving for x, we get:

⇒ x = 283sin(45 degrees)

      = 199.9 feet

Therefore,

The height of the roller coaster is 199.9 feet.

Method 2: Pythagorean theorem

To use this method,

we can use the fact that the length of the roller coaster, 283 feet, is the hypotenuse of the right triangle.

We can thus use the Pythagorean theorem to find the height of the triangle.

Setting up the equation, we have,

283² = x² + h²

where h is the height of the triangle.

Rearranging the equation, we get,

h = √(283² - x²)

Substituting x = 283/√(2) (since the angle between the hypotenuse and the height of the triangle measures 45 degrees), we get,

h = √(283² - (283/√(2))²)

  = 199.9 feet

Therefore, using either method, we can say that the height of the roller coaster is 199.9 feet.

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The number of bacteria present with the given growth rate at t=5 is [tex]N(5) = 7,500 * e^2[/tex] and option x is the correct answer.

An exponential growth pattern is one in which the rate of increase is proportionate to the value of the quantity being measured at any given time. This indicates that the amount by which the quantity increases in each period is a constant proportion of the quantity's present value. Many branches of mathematics and science, such as physics, biology, and finance, utilise exponential growth. Modeling population expansion, the spread of infectious illnesses, the decay of radioactive materials, and the behavior of financial assets are all popular applications.

Given that, the number of bacteria present was 7,500.

The exponential growth is given by the formula:

[tex]N(t) = N(0) * e^{(kt)}[/tex]

Substituting the values N(0) = 7,500 and the growth rate is k = 2/5 we have:

[tex]N(5) = 7,500 * e^{(2/5 * 5)}\\N(5) = 7,500 * e^2[/tex]

Hence, the number of bacteria present at t=5 is [tex]N(5) = 7,500 * e^2[/tex] and option x is the correct answer.

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The mean, variance, and standard deviation for those who have purchased online goods or services are 72, 46.08, and 6.8, respectively.

That 36% of adult internet users have purchased products or services online and for a random sample of 200 adult internet users, we need to find the mean, variance, and standard deviation for the number who have purchased goods or services online.

Therefore, the mean, variance, and standard deviation for the number who have purchased goods or services online are 72, 46.08, and 6.8, respectively.

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Therefore, the solution to the system of equations is: x = 1/5 and y = 8.

An equation is a mathematical statement that asserts the equality of two expressions. Equations contain variables, which are symbols that represent unknown or varying quantities, and constants, which are known values. The variables and constants are connected by mathematical operations, such as addition, subtraction, multiplication, division, and exponentiation, and the resulting expression on both sides of the equation must have the same value. Equations are used to solve problems, find unknowns, and express relationships between quantities. Some common types of equations include linear equations, quadratic equations, polynomial equations, exponential equations, and trigonometric equations.

Here,

We have the following system of equations:

y = 10x + 6

2y + 5x = 17

Using substitution method, we can substitute the expression for y in the second equation with the expression for y in the first equation:

2(10x + 6) + 5x = 17

Simplifying:

20x + 12 + 5x = 17

25x = 5

x = 5/25 = 1/5

Now we can substitute this value of x into either of the two original equations to find the corresponding value of y. Using the first equation:

y = 10(1/5) + 6

y = 2 + 6

y = 8

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If two indistinguishable dice are rolled, what is the probability of the event {(3, 3), (2, 3), (1, 3)}The probability of the event {(3, 3), (2, 3), (1, 3)}  

If two indistinguishable dice are rolled, it is 3/36 or 1/12.

Explanation: Indistinguishable dice are dice that appear identical to one another but do not have unique markings. As a result, indistinguishable dice will have the same number of faces, but the values on each face will be identical.

The total number of possible outcomes is 6 * 6 = 36 because there are six possible outcomes for each roll of a single die.

The probability of rolling the numbers (3, 3), (2, 3), or (1, 3) can be determined as follows: 3/36 or 1/12

For each die, there are six possible outcomes, so there are 6*6, or 36 possible outcomes for two dice.

Because (3, 3), (2, 3), and (1, 3) are the only possible ways to obtain a 3 on one of the dice and a 3, 2, or 1 on the other, the probability is 3/36 or 1/12.

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The lowest temperature ever recorded on earth of -89.2°C is approximately equal to -97°F when rounded to the nearest degree Fahrenheit.

To convert the temperature of -89.2°C to Fahrenheit, we can use the formula:

[tex]F = (C * 1.8) + 32[/tex]

Substituting:

°F = (-89.2 × 1.8) + 32

°F = -128.56 + 32

°F = -97

Therefore, the lowest temperature ever recorded on earth of -89.2°C is approximately equal to -97°F when rounded to the nearest degree Fahrenheit. It's important to note that this is just an approximation as we rounded the result to the nearest degree. However, it gives us a good idea of how extremely cold the temperature was. It's worth mentioning that at such low temperatures, it's important to take appropriate precautions to avoid any adverse health effects.

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Answer:

A questionnaire can collect quantitative, qualitive or both types of data.

Step-by-step explanation:

Answer:

Categorical data

Step-by-step explanation:

Data that relates to certain categories e.g males, females or any types of car

The given parametric equations are X(t) = -3/2 + 2t and y(t) = vt² + 16, the rate of change of y with respect to "t" when t = 3 is 6v

We have the parametric equations that are X(t) = -3/2 + 2t and y(t) = vt² + 16.

At time t, the rate of change of y with respect to t is given by the derivative of y with respect to t, that is dy/dt.

So, y(t) = vt² + 16

Differentiating with respect to t, we get

⇒ dy/dt = 2vt.

Now, t = 3 gives us,

y(3) = v(3)² + 16 ⇒ 9v + 16.

Therefore, the rate of change of y with respect to t at t = 3 is

dy/dt ⇒ 2vt ⇒ 2v(3) ⇒ 6v.

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All parts are define in the below points.

A random sample is a subset of a population in which each individual or element in the population has an equal chance of being selected. It is a sampling method used in statistics and research to minimize bias and increase the generalizability of the findings to the larger population.

a. Hypotheses: Null Hypothesis: The mean DNA concentration of the ocean floor specimens is not significantly different from the healthy concentration (µ = 0.31g/m2). Alternative Hypothesis: The mean DNA concentration of the ocean floor specimens is significantly different from the healthy concentration (µ ≠ 0.31g/m2). Using a two-tailed t-test with alpha = 0.05, we find a p-value of 0.0003, which is less than the significance level. Therefore, we reject the null hypothesis and conclude that the mean DNA concentration of the ocean floor specimens is significantly different from the healthy concentration.

b. We would perform a one-tailed t-test with the alternative hypothesis that the mean DNA concentration is less than 0.31g/m2 if the goal was to determine whether the mean DNA concentration of the floor specimens was lower than what is regarded as a healthy concentration. It would have a p-value of 0.00015.

c. Using the 95% confidence interval approach, we construct a one-sided confidence interval for the mean DNA concentration. If the lower bound of the confidence interval is less than 0.31g/m2, we can conclude that the mean DNA concentration is less than the healthy concentration. The 95% confidence interval for the mean is (0.2457g/m2, 0.3105g/m2), which does not include the healthy concentration of 0.31g/m2. Therefore, we can conclude that the mean DNA concentration of the ocean floor specimens is less than the healthy concentration.

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a). We discover a p-value of 0.0003 using a two-tailed t-test with alpha = 0.05, which is below the significance level.

b). Its p-value would be 0.00015.

c). The safe concentration of [tex]0.31g/m^2[/tex] is not included in the 95% confidence interval for the mean, which is [tex](0.2457g/m^2,\ 0.3105g/m^2)[/tex].

A random sample is a portion of a community in which every person or component has an equal chance of being chosen. In statistics and research, it is a sampling technique used to reduce bias and improve the generalizability of the results to a broader population.

A). An hypothesis is a The null hypothesis states that there is no discernible difference between the mean DNA concentration of the ocean bottom samples and the healthy concentration [tex](\mu=0.31g/m^2)[/tex]. Alternative Hypothesis: The mean DNA concentration of the ocean floor samples differs considerably from the healthy concentration [tex](\mu\neq 0.31g/m^2)[/tex] in a statistically significant way. We discover a p-value of 0.0003 using a two-tailed t-test with alpha = 0.05, which is below the significance level. We therefore reject the null hypothesis and come to the conclusion that the mean DNA concentration of the samples from the ocean bottom differs significantly from that of healthy individuals.

B). If the objective was to determine whether the mean DNA concentration of the floor specimens was lower than what is considered as a healthy concentration, we would conduct a one-tailed t-test with the alternative hypothesis that the mean DNA concentration is less than [tex]0.31g/m^2[/tex]. Its p-value would be 0.00015.

C). We create a one-sided confidence interval for the mean DNA concentration using the 95% confidence interval method. The mean DNA concentrationis less than the healthy concentration if the lower limit of the confidence interval is less than [tex]0.31g/m^2[/tex]. The safe concentration of [tex]0.31g/m^2[/tex] is not included in the 95% confidence interval for the mean, which is [tex](0.2457g/m^2,\ 0.3105g/m^2)[/tex]. As a result, we can say that the average DNA concentration of the samples from the ocean bottom is lower than the healthy concentration.

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The probability that exactly 10 out of 23 cars will not have engine failure is 0.007638.

Step-by-step explanation: First, calculate the probability of an engine failing in five years with no routine maintenance, which is 27.5%. This can be written as a decimal, 0.275.Next, calculate the probability of an engine not failing in five years with routine maintenance. This probability is 100%-27.5% = 72.5%, written as a decimal 0.725.

Now, using the Binomial Distribution formula (nCr), calculate the probability of exactly 10 engines not failing out of 23 cars, where n = 23, r = 10 and p = 0.725. The equation would be [tex](23C10)*(0.725^{10})*(0.275^{13}) = 0.0076379904[/tex]

Finally, round the result to 6 decimal places, giving an answer of 0.007638.

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