All categories

3 years ago

The inequality 5x - 2y 1 > 0 is satisfied by point (-2, -1). true false

Mathematics

2 answers:

alexgriva [62]3 years ago

8 0

Answer: Hello there!

the inequality is (i think):

5x - 2y + 1 > 0

and we wanna see if the point (-2, -1) satisfies this inequality.

If we want to see it, we just need to replace the values x= -2, y = -1 in the inequality and see if it is satisfied:

5x - 2y + 1 > 0

5*(-2) - 2*(-1) + 1 > 0

-10 + 3 > 0

-7 > 0

this is false; then the point (-2, -1) does not satisfy the inequality.

sleet_krkn [62]3 years ago

7 0

2y < 5x + 1
y < 5/2x + 1/2

x = - 0.2
y = 1/2

the point is false

You might be interested in

The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction

Len [333]

Answer:

(a) n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable X represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, μ = $16,642 and standard deviation is, σ = $2,400.

Assuming that the random variable X follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

For a sample size of n = 20

$P(\mu-200$

$=P(-0.37$

For a sample size of n = 50

$P(\mu-200$

$=P(-0.59$

For a sample size of n = 100

$P(\mu-200$

$=P(-0.83$

For a sample size of n = 500

$P(\mu-200$

$=P(-1.86$

n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ( $\bar x$ ) approaches the whole population mean ( $\mu_{x}$ ).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

8 0

3 years ago

Convert to a decimal. Be sure to mark repeating decimals.

Sergio [31]

Answer:

$.24\\is\\reapeating\\0.24$

Step-by-step explanation:

8 0

2 years ago

M=-3,(-2,9) what is this equation in point-slope form? and please explain how

Alik [6]

Answer:

y-9= -3(x+2)

Step-by-step explanation:

y-y1= "M" (x-x1)

y1= 9

X1= -2

5 0

3 years ago

Read 2 more answers

(-8,-2) and (-4,6) write a equation

Dmitry [639]

Answer:

slope of AB=2

AB(distance)=4√5

Step-by-step explanation:

slope of AB=6-(-2)/-4-(-8)

AB=√(-8-(-4))^(2+(-2-6)^(2

=4√5 or 8.94427

next time pls type clearly ur question, because actually I don't know what kind of answer that u want to know. sorry

7 0

3 years ago

Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co

jeka57 [31]

Answer:

<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If $\sum a_n$ is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

$\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho$

If $\rho$ < 1, the series converges absolutely

If $\rho > 1$ , the series diverges

If $\rho = 1$ , the test fails.

Given the series $\sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}$

To test for convergence or divergence using ratio test, we will use the condition above.

$a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}$

$\frac{a_n_+_1}{a_n} = \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\$

$\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\$

aₙ₊₁/aₙ =

$\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\$

note that any constant dividing infinity is equal to zero

$|\frac{1+2/\infty+1/\infty^2}{5}|\\\\$

$\frac{1+0+0}{5}\\ = 1/5$

$\rho = 1/5$

Since The limit of the sequence given is less than 1, hence the series converges.

5 0

3 years ago