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3 years ago

How do you solve the inequality 6y+3> -18x

1 answer:

6 0

$6y + 3 \ \textgreater \ - 18 x$

Subtract 3 from both sides:

$6 y + 3 - 3 \ \textgreater \ - 18 x - 3$

$6y \ \textgreater \ - 18 x - 3$

Dividade both sides by 6 :

$\frac{6y}{6} \ \textgreater \ \frac{-18x-3}{6}$

$y \ \textgreater \ \frac{-18x-3}{6}$

hope this helps!

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There are 4 baskets that store 16 bathroom towels. Consider the number of bathroom towels per basket

WINSTONCH [101]

Answer:

there are 4 bathroom towels per basket

Step-by-step explanation:

the equation is 16/4=4 hope this helps :)

5 0

2 years ago

Read 2 more answers

Marla filled up her car's gas tank and then went on a trip. After she drove 329 miles, she filled her tank with 14 gallons of ga

romanna [79]

Answer:

23 1/2 miles per gallon

Step-by-step explanation:

We start with dividing 329 miles with 14 gallons because it asks to find the number of miles on each gallon of gas -

Miles

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Each gallon of gas

329 ÷ 14 = 23 1/2 miles per gallon

6 0

3 years ago

If the inspection division of a county weights and measures department wants to estimate the mean amount of

Amanda [17]

If inspection department wants to estimate the mean amount with 95% confidence level with standard deviation 0.05 then it needed a sample size of 97.

Given 95% confidence level, standard deviation=0.05.

We know that margin of error is the range of values below and above the sample statistic in a confidence interval.

We assume that the values follow normal distribution. Normal distribution is a probability that is symmetric about the mean showing the data near the mean are more frequent in occurence than data far from mean.

We know that margin of error for a confidence interval is given by:

Me= $z/2* ST/\sqrt{N}$

α=1-0.95=0.05

α/2=0.025

z with α/2=1.96 (using normal distribution table)

Solving for n using formula of margin of error.

$n=(z/2ST/Me)^{2}$

n= $(1.96*0.05)^{2} /(0.01)^{2}$

=96.4

By rounding off we will get 97.

Hence the sample size required will be 97.

Learn more about standard deviation at brainly.com/question/475676

#SPJ4

The given question is incomplete and the full question is as under:

If the inspection division of a county weights and measures department wants to estimate the mean amount of soft drink fill in 2 liters bottles to within (0.01 liter with 95% confidence and also assumes that standard deviation is 0.05 liter. What is the sample size needed?

7 0

2 years ago

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

3 years ago

5+3(q-4)=2(q+1) I don’t understand how to solve this question

Shtirlitz [24]

Answer:

q = 9

Step-by-step explanation:

5 + 3(q - 4) = 2(q + 1)

To solve, you need to get q by itself on one side.

5 + 3q - 12 = 2q + 2 --- distribute the multipliers next to the parenthesis

3q - 7 = 2q + 2 --- combine like terms

q - 7 = 2 --- subtract 2q from both sides

q = 9 --- add 7 to both sides

3 0

3 years ago