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aleksandr82 [10.1K]
3 years ago
8

Evaluate |x|+2.7​ for x=−1.5​ .

Mathematics
1 answer:
Olenka [21]3 years ago
3 0

Answer:

1.2

Step-by-step explanation: Hope it helps

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Answer: C

Step-by-step explanation:

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Solve the equation. f - 8/9 = 5/9<br> f = ?
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f=13/9

Step-by-step explanation:

f-8/9=5/9

solution

f=5/9+8/9

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Read 2 more answers
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
Zane way to color meatballs during a science class the blue ball weighs 7 lb in the green ball weighed 0.69 lb in the same place
Umnica [9.8K]

you would just add both of them together so the weight would be 7.69 Ibs

7 0
3 years ago
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