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MA_775_DIABLO [31]

3 years ago

13

HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP ASAP ASAP ASAP ASAP ASAP ASAP ASAP ASAP ASAP ASAP ASAP ASAP ASAP ASAP ASA

P ASAP HELP

1 answer:

xxMikexx [17]3 years ago

6 0

Ok I know how to do this but what do I need to do?

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Can u help me on this pls?

GenaCL600 [577]

Answer:

The table:

x | y

2 5

3 7.5

6 15

Explanation:

Yes, Y is proportional to X and Y because it follows the equation which is

y = 5/2x

8 0

2 years ago

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For each expression, combine like terms and write an equivalent expression with fever terms.

ASHA 777 [7]

Answer:

1. 7x

2. 8x - 1

3. 6x + 12

4. 3x + 4

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Step-by-Step Explanation:

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2 years ago

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What’s the answer to -3 + 47 = 5x + 103

Solnce55 [7]

Answer:

-11.8

Step-by-step explanation:

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3 years ago

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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 - <em>α</em>)% 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 <em>z</em> 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

Classify the following triangle. Check all that apply

kramer

D and E. If I remember right that should be the answer.

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3 years ago

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