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3 years ago

Anyone Know This Answer ?

Mathematics

2 answers:

OverLord2011 [107]3 years ago

7 0

I hope this helps you

M^2+R^2=W^2

M^2+16=25

M^2=9

Nostrana [21]3 years ago

5 0

I'm not 100% positive, but I believe it's 41. Please feel free to correct me on this though

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What is the average rate of change of the function over the interval x = 0 to x = 8?

salantis [7]

Ans: The average rate of change of function = $\frac{1}{102}$ (in fraction)

Explanation:
The average rate of change of function = $\frac{f(b) - f(a)}{b - a}$

Where
b = 8 (x's upper bound)
a = 0 (x's lower bound)

$f(b) = f(8) = \frac{2*8-2}{5*8-6} = \frac{7}{17} \\ f(a) = f(0) = \frac{2*0-2}{5*0-6} = \frac{1}{3} \\ \frac{f(b) - f(a)}{b-a} = \frac{ \frac{7}{17} - \frac{1}{3}}{8} = \frac{1}{102}$

Hence the average rate of change of function = $\frac{1}{102}$ (in fraction)

5 0

3 years ago

Y is greater than or equal to -×-5

Katen [24]

Answer:

Y>-x-5

Step-by-step explanation:

6 0

3 years ago

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

Eight times what equals 226

PSYCHO15rus [73]

Answer:

28.25

Step-by-step explanation:

just do 226 divided by 8

3 0

3 years ago

Read 2 more answers

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