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ale4655 [162]
3 years ago
7

Simplify the expression: - 4t+ - 7t+3–10t

Mathematics
1 answer:
Gekata [30.6K]3 years ago
8 0

Answer:t+3

Step-by-step explanation:

4t+7t+3-10t

11t+3-10t

t+3

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4mn,m and, 5. The terms are anything that's not a symbol like + or = in an equation

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There are 108 inches in 9 feet

Step-by-step explanation:

Well, two unit multipliers mean we have to convert through two units before we arrive at our final answer so we can convert from feet to yards to inches!

We know that 3 feet equal a yard, right?

So we can divide 9 by 3 to get the amount that 9 feet translates to in yards.

9/3 = 3!

So we are now done with our first conversion! Our second is going to be from yards to inches!

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2, 10 after a dilation by a scale factor of 1/2 centered at the<br> origin?
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The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses.
Luden [163]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

If we had increased the confidence level, the margin of error also would have increased.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96. Increasing the confidence level, z also increases, hence the margin of error also would have increased.

The sample size and the estimate are given as follows:

n = 450, \pi = 0.47.

The lower and the upper bound of the interval are given, respectively, by:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.4239

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.5161

The 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

5 0
1 year ago
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