What is the derivative of y=ln(ln(x))?

Identify the type of function represented by f(x)=1/2•(3)^x

dalvyx [7]

ANSWER

An exponential growth function.

EXPLANATION

The given function is

$f(x) = \frac{1}{2} {(3})^{x}$

This function is of the form,

$f(x) = a {(b})^{kx}$

Since k is positive , the function will grow exponentially and fall asymptote to the negative x-axis.

As x-values becomes bigger and bigger positively, the graph grows exponentially.

6 0

3 years ago

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Ten college students were randomly selected. Their grade point averages (GPAs) when they entered the program were between 3.5 a

solong [7]

Answer:

The 95% confidence interval for the true slope is (0.03985, 0.14206).

Step-by-step explanation:

For the regression equation:

$\hat y=\alpha +\hat \beta x$

The (1 - <em>α</em>)% confidence interval for the regression coefficient or slope $(\hat \beta )$ is:

$Ci=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )$

The regression equation for current GPA (Y) of students based on their GPA's when entering the program (X) is:

$\hat Y=3.584756+0.090953 X$

The summary of the regression analysis is:

Predictor Coefficient SE t-stat p-value

Constant 3.584756 0.078183 45.85075 5.66 x 10⁻¹¹

Entering GPA 0.090953 0.022162 4.103932 0.003419

The regression coefficient and standard error are:

$\hat \beta = 0.090953\\SE (\hat \beta)=0.022162$

The critical value of <em>t</em> for 95% confidence level and 8 degrees of freedom is:

$t_{\alpha/2, n-2}=t_{0.05/2, 10-2}=t_{0.025, 8}=2.306$

Compute the 95% confidence interval for $(\hat \beta )$ as follows:

$CI=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )\\=0.090953\pm 2.306\times 0.022162\\=0.090953\pm 0.051105572\\=(0.039847428, 0.142058572)\\\approx (0.03985, 0.14206)$

Thus, the 95% confidence interval for the true slope is (0.03985, 0.14206).

6 0

3 years ago

ASAP

s2008m [1.1K]

Answer:

4

Step-by-step explanation:

2×4 is the only one that equals less than 9, 2× 4.5 is 9, 2×5.5 is 11, 2×6 is 12, and the question is trying to get below 9, so 4 is the only answer that can get below 9

8 0

3 years ago

A filter filled with liquid is in the shape of a vertex-down cone with a height of 9 inches and a diameter of 6 inches at its op

Alina [70]

Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

Step-by-step explanation:

Since we have given that

Height = 9 inches

Diameter = 6 inches

Radius = 3 inches

So, $\dfrac{r}{h}=\dfrac{3}{9}=\dfrac{1}{3}\\\\r=\dfrac{1}{3}h$

Volume of cone is given by

$V=\dfrac{1}{3}\pi r^2h\\\\V=\dfrac{1}{3}\pi \dfrac{1}{9}h^2\times h\\\\V=\dfrac{1}{27}\pi h^3$

By differentiating with respect to time t, we get that

$\dfrac{dv}{dt}=\dfrac{1}{27}\pi \times 3\times h^2\dfrac{dh}{dt}=\dfrac{1}{9}\pi h^2\dfrac{dh}{dt}$

Now, the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie $\dfrac{dv}{dt}=-4\ in^3$

and h = 2 inches deep.

$-4=\dfrac{1}{9}\times \pi\times (2)^2\dfrac{dh}{dt}\\\\-9\pi =\dfrac{dh}{dt}\\\\-28.28=\dfrac{dh}{dt}$

Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

7 0

3 years ago

Pls find x, has to do with trig and answer has to be in radical form

MrMuchimi

I don’t have a calculator with me use sine 20divided by Sin(60) will give you x

7 0

3 years ago

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