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avanturin [10]
3 years ago
13

Let t represent an unknown number. What is the value

Mathematics
1 answer:
bekas [8.4K]3 years ago
4 0

There are two possible answers:

t = 2/3   or   t = -1/3

========================================================

Explanation:

The expression x+3 is the same as x-(-3). It is of the form x-k

Recall that by the remainder theorem, we know that p(x) is divisible by (x-k) if and only if p(k) = 0.

In this case, k = -3. Plug this into the p(x) function and we get

p(x) = t^2*x^3 - 3t*x + 6

p(-3) = t^2*(-3)^3 - 3t*(-3) + 6

p(-3) = -27t^2 + 9t + 6

Set that equal to 0

p(-3) = 0

-27t^2 + 9t + 6 = 0

-3(9t^2 - 3t - 2) = 0

9t^2 - 3t - 2 = 0

Now apply the quadratic formula to solve for t

t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\t = \frac{-(-3)\pm\sqrt{(-3)^2-4(9)(-2)}}{2(9)}\\\\t = \frac{3\pm\sqrt{81}}{18}\\\\t = \frac{3\pm9}{18}\\\\t = \frac{3+9}{18}\ \text{ or } \ t = \frac{3-9}{18}\\\\t = \frac{12}{18}\ \text{ or } \ t = \frac{-6}{18}\\\\t = \frac{2}{3}\ \text{ or } \ t = -\frac{1}{3}\\\\

We get two possible solutions for t.

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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
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Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

n= 955 represent the sampel size slected

x = 812 number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

The confidence interval for the proportion  would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the significance is \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution and we got.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

8 0
3 years ago
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