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Rom4ik [11]
3 years ago
14

How many integers n satisfy the inequality 3n² - 4≤44?

Mathematics
1 answer:
Margaret [11]3 years ago
7 0

Answer:

The possible values of n  for n ≤ +4  =  (-∞,+4]

The possible values of n for  n ≤ -4  =  (-∞,-4]

Step-by-step explanation:

Here, the given inequality is:

3n^2 - 4 \leq 44

Firstly, let us solve the given inequality for the desirable value of n.

3n^2 - 4 \leq 44

Adding 4 on both sides, we get:

3n^2 - 4 + 4 \leq 44 + 4

3n^2  \leq 48

or, n^2 \leq \frac{48}{3}\implies n^2 \leq 16\\

⇒   n ≤ +4  or  n ≤ -4

So, the possible values of n  for  n ≤ +4  =  (-∞,+4]

And,  the possible values of n  for  n ≤ -4  =  (-∞,-4]

So, we can pick any of the integer values from the both defined sets.

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X= -1. Y=-5

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The x decreases by -1 and the y decreases by -5 .

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Use the Distributive Property to solve the equation.
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Answer:

x = -6

Step-by-step explanation:

-5(x + 3) = 15

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-5x -15 =15

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Which equation represents a linear function that has a slope of 4/5<br> and a y-intercept of –6?
vredina [299]

Answer:

\large\boxed{y=\dfrac{4}{5}x-6}

Step-by-step explanation:

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y=mx+b

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Read 2 more answers
Health insurers are beginning to offer telemedicine services online that replace the common office visit. A company provides a v
Veronika [31]

Answer:

\text {CI} = (60.54, \: 81.46)\\\\

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

Step-by-step explanation:

Let us find out the mean savings for a televisit to the doctor from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

\bar{x} = \$71  

Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

 s = \$ 22.35

The confidence interval is given by

\text {confidence interval} = \bar{x} \pm MoE\\\\

Where the margin of error is given by

$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and t_{\alpha/2} is the t-score corresponding to a 95% confidence level.

The t-score is given by is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 20 - 1 = 19

From the t-table at α = 0.025 and DoF = 19

t-score = 2.093

So, the margin of error is

MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\MoE = 2.093\cdot \frac{22.35}{\sqrt{20} } \\\\MoE = 2.093\cdot 4.997\\\\MoE = 10.46\\\\

So the required 95% confidence interval is

\text {CI} = \bar{x} \pm MoE\\\\\text {CI} = 71 \pm 10.46\\\\\text {CI} = 71 - 10.46, \: 71 + 10.46\\\\\text {CI} = (60.54, \: 81.46)\\\\

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

7 0
3 years ago
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