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3 years ago

Which of the m-values satisfy the following inequality?

Mathematics

1 answer:

dangina [55]3 years ago

5 0

Answer:

a m=0

Step-by-step explanation:

5(0)+1 = 0+1 = 1

1<4

hope this helps

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The inequality x + 2y ≥ 3 is satisfied by point (1, 1). True False

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In the given statement above, in this case, the answer would be TRUE. It is true that the inequality x + 2y ≥ 3 is satisfied by point (1, 1). In order to prove this, we just have to plug in the values. 1 + 2(1) ≥ 3
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A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

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3 years ago

Solve for x in the equation X2-8x+41 = 0 -

djverab [1.8K]

Answer:

The value of x fro the given equation is ( 4 + 2 i ) , ( 4 - 2 i )

I.e option D

Step-by-step explanation:

Given equation as :

x² - 8 x + 41 = 0

For quadratic equation ax² + b x + c = 0

The value of x = $\frac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}$