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Which of the following is the equation of the line that is parallel toy= 3/5x+ 8 and goes through point (-10,4)?
Select one:
a. y = 5/3x + 20 2/3
b. y=-5/3x – 12 2/3
c.y= 3/5x + 10
d. y = -3/5x-2

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

We want to write the equation of a line that is parallel to:

$y=(3)/(5)x+8$

And also passes through (-10, 4).

Remember that parallel lines have the same slope.

The slope of our old line is 3/5.

Therefore, the slope of our new line is also 3/5.

We know that it passes through (-10, 4). So, we can use the point-slope form:

$y-y_1=m(x-x_1)$

Where m is the slope and (x₁, y₁) is a point.

So, let's substitute 3/5 for m and let (-10, 4) be our (x₁, y₁). This yields:

$y-(4)=(3)/(5)(x-(-10))$

Simplify:

$y-(4)=(3)/(5)(x+10)$

Distribute on the right:

$y-4=(3)/(5)x+6$

Add 4 to both sides:

$y=(3)/(5)x+10$

So, our answer is C.

And we're done!

Answer 2

Answer:

Step-by-step explanation:

Hey there!

The equation of a st.line passing through point (-10,4) is ;

(y-y1)= m1(x-x1) [one point formula]

Put all values.

(y - 4) = m1( x + 10)..........(i)

Another equation is; y = 3/5 + 8.............(ii)

From equation (ii)

Slope (m2) = 3/5 [ By comparing equation with y = mx+c].

As per the condition of parallel lines,

Slope of equation (i) = slope of equation (ii)

(i.e m1 = m2 )

Therefore, the value of m1 is 3/5.

Putting value of slope in equation (i).

$(y - 4) = (3)/(5) (x + 10)$

$(y - 4) = (3)/(5) x + (3)/(5) * 10$

$(y - 4) = (3)/(5) x + 6$

$y = (3)/(5) x + 10$

Therefore the required equation is y = 3/5x + 10.

Hopeit helps...

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An = −7.1 − 2.1n Find a27

Answers

The 27th term of the Arithmetic sequence defined by an = -7.1 -2.1n is -63.6. This is determined by substituting n=27 into the formula.

The problem here is asking us to find the 27th term, denoted as a27, of an arithmetic sequence. The general formula for the nth term of such a sequence is an = a1 + (n-1) * d, where a1 is the first term and d is the common difference.

However, in this case, the formula given is an = −7.1 − 2.1n. Therefore, we just need to substitute n = 27 into the formula to find a27.

Following that, we have a27 = -7.1 -2.1*27. Doing the calculation, we get a27 = -63.6. Therefore, the 27th term of the sequence is -63.6.

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Answer:

An = -63.8

Step-by-step explanation:

−7.1 − 2.1n

−7.1 − 2.1(27)

-7.1 - 56.7 = -63.8

An = -63.8

If x1, x2, . . . , xn are independent and identically distributed random variables having uniform distributions over (0, 1), find (a) e[max(x1, . . . , xn)]; (b) e[min(x1, . . . , xn)].

Answers

Denote by $X_((n))$ the maximum order statistic, with $X_((n))=\max\{X_1,\ldots,X_n\}$ , and similarly denote by $X_((1))$ the minimum order statistic. Then the CDF for $X_((n))$ is

$F_{X_((n))}(x)=\mathbb P(X_((n))\le x)$

In order for there to be some $x$ that exceeds the value of $X_((n))$ , it must be true that $x$ exceeds the value of all the $X_i$ , so the above is equivalent to the joint probability

$F_{X_((n))}(x)=\mathbb P(X_1\le x,\ldots,X_n\le x)$

and since the $X_i$ are i.i.d., we have

$F_{X_((n))}(x)=\mathbb P(X_1\le x)\cdots\mathbb P(X_n\le x)=\mathbb P(X_1\le x)^n$
$\implies F_{X_((n))}(x)=F_X(x)^n$

where $X\sim\mathrm{Unif}(0,1)$ . We have

$F_X(x)=\begin{cases}0&\text{for }x<0\nx&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$

and so

$F_{X_((n))}(x)=\begin{cases}0&\text{for }x<0\nx^n&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$
$\implies f_{X_((n))}(x)=\begin{cases}nx^(n-1)&\text{for }0<x<1\n0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_((n))]=\displaystyle\int_0^1xnx^(n-1)\,\mathrm dx=n\int_0^1x^n\,\mathrm dx=\frac n{n+1}$

Using similar reasoning, we can find the CDF for $X_((1))$ . We have

$F_{X_((1))}(x)=\mathbb P(X_((1))\le x)=1-\mathbb P(X_((1))>x)$
$F_{X_((1))}(x)=1-\mathbb P(X_1>x,\ldots,X_n>x)=1-\mathbb P(X_1>x)^n$
$F_{X_((1))}(x)=1-(1-\mathbb P(X\le x))^n=1-(1-F_X(x))^n$
$\implies F_{X_((1))}(x)=\begin{cases}0&\text{for }x<0\n1-(1-x)^n&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$
$\implies f_{X_((1))}(x)=\begin{cases}n(1-x)^(n-1)&\text{for }0<x<1\n0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_((1))]=\displaystyle\int_0^1xn(1-x)^(n-1)\,\mathrm dx=\frac1{n+1}$

Final answer:

The expected values of the maximum and minimum of independent and identically distributed (iid) uniform random variables, x1, x2, ..., xn, are given by E[max(x1, ..., xn)] = n / (n + 1) and E[min(x1, ..., xn)] = 1 / (n + 1) respectively.

Explanation:

In mathematics, particularly in probability theory and statistics, the question is related to independent and identically distributed (iid) random variables with a uniform distribution. The expected value or mean (E) of the maximum (max) and minimum (min) of these random variables is sought.

(a) The expected value of the max of 'n' iid uniform random variables, x1, x2, ..., xn, is calculated by integrating the nth power of x from 0 to 1. It can be found via the equation E[max(x1, ..., xn)] = n / (n + 1).

(b) Similarly, the expected value of the min of 'n' iid uniform random variables is acquired by doing (1 / (n + 1)). Hence, E[min(x1, ..., xn)] = 1 / (n + 1).

By understanding these, you could visualize the various outcomes of the random variables and their distributions, demonstrating how likely each outcome could occur.

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Find the Greatest common factor of the numbers using lists of factors

8,12

Answers

The answer is 4 bc you can multiply 4 with both 8 and 12. I’m sorry but I’m bad at explaining but the answer is 4 :(

The integral $\pi\int\limits^2_1 {(1-(lnx)^(2)) } \, dx$ represents the volume of a solid obtained by rotating a region around y=-1. Evaluate.

Answers

Answer:

V = π (-2 (ln 2)² + 4 ln 2 − 1)

V ≈ 2.55

Step-by-step explanation:

V = π ∫₁² (1 − (ln x)²) dx

V/π = ∫₁² (1 − (ln x)²) dx

V/π = ∫₁² dx − ∫₁² (ln x)² dx

V/π = x |₁² − ∫₁² (ln x)² dx

V/π = 1 − ∫₁² (ln x)² dx

To evaluate the second integral, integrate by parts.

If u = (ln x)², then du = 2 (ln x) / x dx.

If dv = dx, then v = x.

∫ u dv = uv − ∫ v du

= (ln x)² x − ∫ x (2 (ln x) / x) dx

= x (ln x)² − 2 ∫ ln x dx

Integrate by parts again.

If u = ln x, then du = 1/x dx.

If dv = dx, then v = x.

∫ u dv = uv − ∫ v du

= x ln x − ∫ x (1/x dx)

= x ln x − ∫ dx

= x ln x − x

Substitute:

∫ (ln x)² dx = x (ln x)² − 2 ∫ ln x dx

∫ (ln x)² dx = x (ln x)² − 2 (x ln x − x)

∫ (ln x)² dx = x (ln x)² − 2x ln x + 2x

Substitute again:

V/π = 1 − ∫₁² (ln x)² dx

V/π = 1 − (x (ln x)² − 2x ln x + 2x) |₁²

V/π = 1 + (-x (ln x)² + 2x ln x − 2x) |₁²

V/π = 1 + (-2 (ln 2)² + 4 ln 2 − 4) − (-1 (ln 1)² + 2 ln 1 − 2)

V/π = 1 − 2 (ln 2)² + 4 ln 2 − 4 + 2

V/π = -2 (ln 2)² + 4 ln 2 − 1

V = π (-2 (ln 2)² + 4 ln 2 − 1)

V ≈ 2.55

The line AB has midpoint (2,5).
A has coordinates (1, 2).
Find the coordinates of B.

Answers

Answer:

$X_m = (A_x +B_x)/(2)= (1+B_x)/(2)= 2$

And we can solve for $B_x$ and we got:

$1+B_x = 4$

$B_x = 3$

$Y_m = (A_y +B_y)/(2)= (2+B_y)/(2)= 5$

And we can solve for $B_x$ and we got:

$2+B_y = 10$

$B_y = 8$

So then the coordinates for B are (3,8)

Step-by-step explanation:

For this case we know that the midpoint for the segment AB is (2,5)

And we know that the coordinates of A are (1,2)

We know that for a given segment the formulas in order to find the midpoint are given by:

$X_m = (A_x +B_x)/(2)= (1+B_x)/(2)= 2$

And we can solve for $B_x$ and we got:

$1+B_x = 4$

$B_x = 3$

$Y_m = (A_y +B_y)/(2)= (2+B_y)/(2)= 5$

And we can solve for $B_x$ and we got:

$2+B_y = 10$

$B_y = 8$

So then the coordinates for B are (3,8)

Final answer:

The coordinates of point B are (4, 8).

Explanation:

To find the coordinates of point B, we can use the midpoint formula. The midpoint formula states that the coordinates of the midpoint M(xm, ym) of a line segment with endpoints A(x1, y1) and B(x2, y2) are given by:

xm = (x1 + x2) / 2

ym = (y1 + y2) / 2

In this case, we are given that the midpoint M is (2, 5) and A is (1, 2). We can substitute these values into the formula:

2 = (1 + x2) / 2

5 = (2 + y2) / 2

Now, we can solve for x2 and y2:

x2 = 4

y2 = 8

Therefore, the coordinates of point B are (4, 8).

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Business Week conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume that the mean annual salary for male and female graduates 10 years after graduation is $168,000 and $117,000, respectively. Assume the standard deviation for the male graduates is $40,000 and for the female graduates it is $25,000. 1. In which of the preceding two cases, part a or part b, do we have a higher probability of obtaining a smaple estimate within $10,000 of the population mean? why? 2. What is the probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean?

Answers

Answer:

1. Due to the lower standard deviation, it is more likely to obtain a sample of females within $10,000 of the population mean

2. 15.87% probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = (X - \mu)/(\sigma)$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = (\sigma)/(√(n))$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

1. In which of the preceding two cases, part a or part b, do we have a higher probability of obtaining a smaple estimate within $10,000 of the population mean? why?

The lower the standard deviation, the less dispersed the values are, meaning it is more likely to find values within a certain threshold of the mean.

Due to the lower standard deviation, it is more likely to obtain a sample of females within $10,000 of the population mean.

2. What is the probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean?

We have that:

$\mu = 168000, \sigma = 40000, n = 100, s = (40000)/(√(100)) = 4000$

This probability is the pvalue of Z when X = 168000 - 4000 = 164000. So

$Z = (X - \mu)/(\sigma)$

By the Central Limit Theorem

$Z = (X - \mu)/(s)$

$Z = (164000 - 168000)/(4000)$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

15.87% probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean

Final answer:

1. We have a higher probability of obtaining a sample estimate within $10,000 of the population mean when the standard deviation is smaller. In this case, the standard deviation for female graduates is smaller, so the probability is higher. 2. The probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean can be calculated using the z-score formula and the z-table.

Explanation:

1. In the case where the standard deviation is smaller, we have a higher probability of obtaining a sample estimate within $10,000 of the population mean. This is because a smaller standard deviation indicates less variability in the data, making it more likely for the sample mean to be closer to the population mean. In this case, the standard deviation for female graduates is smaller, so the probability is higher.

2. To calculate the probability, we need to calculate the z-score and then use the z-table. The z-score formula is z = (x - μ) / (σ / sqrt(n)), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the given values, we find the z-score and use the z-table to find the probability.

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