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4 years ago

A light bulb consumes

2 answers:

7 0

For this we first need to know how much watt hours are used in 12 house which we can find by taking 960/2=480 now we just need to take 960*4=3840 and add 480 which gives us are end answer of 4320 watt hours. Enjoy!=)

Readme [11.4K]4 years ago

4 0

The answer would be 4,320 watts because 960*4.5=4,320. (The 4.5 came from the amount of days because 12 hours is 1/2 one day)

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6x+18y-(53w+x-y+29) - (6x-(1+11)<br> x=2 y=3 w=-3

Luba_88 [7]

12+54+131 - (12-(12))
197 -(12-12)
197-0
197

7 0

3 years ago

Read 2 more answers

What's the circumference of a circle if its 12m long

Oliga [24]

Answer:

75.4

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A particular psychological test is used to measure need for achievement. The average test score for all university students in O

Angelina_Jolie [31]

Answer:

Only B and C are always true.

Step-by-step explanation:

To analyse which statements are always true, we go through the process of finding confidence intervals for sample means, from the start.

Confidence Interval = (Sample mean) ± (Margin of error)

From this expression, it is evident that the margin of error determines how wide the confidence interval would be.

Sample Mean = 110 (given)

Margin of Error = (Critical value) × (Standard deviation of the distribution of sample means)

Since no information about the population standard deviation is provided, the critical value is obtained using t-distribution.

The critical value usually varies at different confidence levels and degree of freedoms.

The higher the confidence level, the higher the critical value and the higher the margin of error leading to a wider range.

Hence, a confidence interval of 95% will have a higher critical value than a confidence interval of 90%. Hence, statement C is proved once that 'for n = 100, the 95% confidence interval will be wider than the 90% confidence interval'.

After obtaining the critical value, we then obtain the standard deviation of the distribution of sample means or simply the standard error of the mean. This is given as

σₓ = σ/√n

where σ = standard deviation; which isn't given. The standard deviation might be high enough to guarantee that the Margin of error is high too for the confidence interval to contain 115 or low enough to ensure that the Margin of error is very small and the confidence interval will not contain 115.

Or the sample size might be high enough to make the standard error of the mean to be eventually small and lead to a small margin of error and the condidence interval will not contain 115.

The point is, it isn't always sure that the resulting interval.would contain 115. So, statement A isn't always true.

Then from σₓ = σ/√n,

n = sample size, a large sample size means a more narrow confidence interval and a small sample size means a wider sample size. This proves statement C.

The 95% confidence interval for n = 100 will be more narrow than the 95% confidence interval for n = 50.

Hence, Only B and C are always true.

Hope this Helps!!!

5 0

3 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

mote1985 [20]

Answer:

$\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

Step-by-step explanation:

To find the derivative of the function $y(x)=\ln \left(\frac{x}{x^2+1}\right)$ you must:

Step 1. Rewrite the logarithm:

$\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 2. The derivative of a sum is the sum of derivatives:

$\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)$

Step 3. The derivative of natural logarithm is $\left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}$

${\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 4. The function $\ln{\left(x^{2} + 1 \right)}$ is the composition $f\left(g\left(x\right)\right)$ of two functions $f\left(u\right)=\ln{\left(u \right)}$ and $u=g\left(x\right)=x^{2} + 1$

Step 5. Apply the chain rule $\left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }$

$-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}$

Return to the old variable:

$- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}$

The derivative of a sum is the sum of derivatives:

$- \frac{{\frac{d}{dx}\left(x^{2} + 1\right)}}{x^{2} + 1} + \frac{1}{x}=- \frac{{\left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right)}}{x^{2} + 1} + \frac{1}{x}=\frac{1}{x^{3} + x} \left(x^{2} - x \left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right) + 1\right)$

Step 6. Apply the power rule $\frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}$

$\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\$

$\frac{1}{x^{3} + x} \left(- x^{2} - x {\frac{d}{dx}\left(1\right)} + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x {\left(0\right)} + 1\right)=\\\\\frac{1 - x^{2}}{x \left(x^{2} + 1\right)}$

Thus, $\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

3 0

3 years ago

PLEASE HELP ME WITH THIS QUESTION!! Thanks

Gre4nikov [31]

Answer:

20% probability

Step-by-step explanation:

First, we need to calculate the area of all the figures inside the rectangle.

Area of triangle= 1/2 BH

So, 3x2=(6)

Area of Square= L^2

3x3=(9)

Area of rectangle= LxW

=60.

60+9+6= 75.

75 is the total area.

Now, lets find the added area of only the triangle and square.

9+6= 15

The probability will be 15/75=

Convert into percent form

= 20%

Hope this helps and please mark me brainliest :)

7 0

2 years ago