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3 years ago

What is the decimal equivalent of 9/32

Mathematics

1 answer:

Charra [1.4K]3 years ago

8 0

.28125

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You intend to estimate a population proportion with a confidence interval. The data suggests that the normal distribution is a r

SIZIF [17.4K]

Answer:

The critical value that corresponds to a confidence level of 99% is, 2.58.

Step-by-step explanation:

Consider a random variable X that follows a Binomial distribution with parameters, sample size n and probability of success p.

It is provided that the distribution of proportion of random variable X, $\hat p$ , can be approximated by the Normal distribution.

The mean of the distribution of proportion is, $\mu_{\hat p}=\hat p$

The standard deviation of the distribution of proportion is, $\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$ .

Then the confidence interval for the population proportion p is:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }$

The confidence level is 99%.

The significance level is:

$\alpha =1-\frac{Confidence\ level}{100}=1-\frac{99}{100}=1-0.99=0.01$

Compute the critical value as follows:

$z_{\alpha /2}=z_{0.01/2}=z_{0.005}$

That is:

$P(Z>z)=0.005\\P(Z$

Use the z-table for the z-value.

For z = 2.58 the P (Z < z) = 0.995.

And for z = -2.58 the P (Z > z) = 0.005.

Thus, the critical value is, 2.58.

7 0

4 years ago

Dilate triangle ABC using a scale factor of 1/5. what are the corresponding coordinates for vertices A', B', and C'?

Natasha_Volkova [10]

Answer:

A' is 1, -3/5

B' is -1, -3/5

C' id -1/5, 1

3 0

3 years ago

Linear Transformations To reflect a linear function about the y-axis, the slope is what

baherus [9]

Answer: Multiplied by -1

Example:

Let's say we had the linear function y = 2x+5

Reflecting this over the y axis means we replace every x with -x and simplify

y = 2x+5 becomes y = 2(-x)+5 and that simplifies to y = -2x+5

The y intercept stays the same the whole time. This is because the y intercept is on the mirror line of the y axis. Stuff on the mirror line does not move.

Notice the jump from the old slope (2) to the new slope (-2) has us multiply by -1.

4 0

3 years ago

Triangle ABC has verrocktes A (2,-1) B ( -3,0) and C ( -1,4) . Find the vertices of the image of triangle and after a translatio

belka [17]

Answer:

A(2,1) , B(-3,2) , C(-1,6)

Step-by-step explanation:

A translation of 2 units up means the y values will be shifted up 2 units

A (2,-1) -> A(2,1)

B (-3,0) -> B(-3,2)

C (-1,4) -> C(-1,6)

3 0

3 years ago

Consider the two functions:

koban [17]

Answer:

a) The x value of the point where the two equations intersect in terms of a is $x=\frac{40}{4+5a}$

b) The value of the functions at the point where they intersect is $\frac{10 (28 + 15 a)}{4 + 5 a}$

c) The partial derivative of f with respect to $x$ is $\frac{\partial f}{\partial x} = -5a$ and the partial derivative of f with respect to $a$ is $\frac{\partial f}{\partial x} = -5x$

d) The value of $\frac{\partial f}{\partial x}(3,2) = -10$ and $\frac{\partial f}{\partial a}(3,2) = -15$

e) $\upsilon_1=-\frac{3}{4} = -0.75$ and $\upsilon_2=-\frac{3}{4} = -0.75$

f) equation $\upsilon_1 = \frac{-5a\cdot x}{70-5ax}=\frac{ax}{ax-14}$ and $\upsilon_2 = \frac{-5a\cdot a}{70-5ax}=\frac{a^2}{ax-14}$

Step-by-step explanation:

a) In order to find the $x$ we just need to equal the equations and solve for $x$ :

$f(x,a)=g(x)\\70-5xa = 30+4x\\70-30 = 4x+5xa\\40 = x(4+5a)\\\boxed {x = \frac{40}{4+5a}}$

b) Since we need to find the value of the function in the intersection point we just need to substitute the result from a) in one of the functions. As a sanity check , I will do it in both and the value (in terms of $a$ ) must be the same.

$f(x,a)=70-5ax\\f(\frac{40}{4+5a}, a) = 70-5\cdot a \cdot \frac{40}{4+5a}\\f(\frac{40}{4+5a}, a) = 70 - \frac{200a}{4+5a}\\f(\frac{40}{4+5a}, a) = \frac{70(4+5a) -200a}{4+5a}\\f(\frac{40}{4+5a}, a) =\frac{280+350a-200a}{4+5a}\\\boxed{ f(\frac{40}{4+5a}, a) =\frac{10(28+15a)}{4+5a}}$

and for $g(x)$ :

$g(x)=30+4x\\g(\frac{40}{4+5a})=30+4\cdot \frac{40}{4+5a}\\g(\frac{40}{4+5a})=\frac{30(4+5a)+80}{4+5a}\\g(\frac{40}{4+5a})=\frac{120+150a+80}{4+5a}\\\boxed {g(\frac{40}{4+5a})=\frac{10(28+15a)}{4+5a}}$

c) $\frac{\partial f}{\partial x} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial x}=0-5a\\\frac{\partial f}{\partial x} =-5a$

$\frac{\partial f}{\partial a} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial a}=0-5x\\\frac{\partial f}{\partial a} =-5x$

d) Then evaluating:

$\frac{\partial f}{\partial x} =-5a\\\frac{\partial f}{\partial x} =-5\cdot 2=-10$

$\frac{\partial f}{\partial a} =-5x\\\frac{\partial f}{\partial a} =-5\cdot 3=-15$

e) Substituting the corresponding values:

$\upsilon_1 = \frac{\partial f(3,2)}{\partial x}\cdot \frac{3}{f(3,2)} \\\upsilon_1 = -10 \cdot \frac{3}{40} = -\frac{3}{4} = -0.75$

$\upsilon_2 = \frac{\partial f(3,2)}{\partial a}\cdot \frac{3}{f(3,2)} \\\upsilon_2 = -15 \cdot \frac{2}{40} = -\frac{3}{4} = -0.75$

f) Writing the equations:

$\upsilon_1=\frac{\partial f (x,a)}{\partial x}\cdot \frac{x}{f(x,a)}\\\upsilon_1=-5a\cdot \frac{x}{70-5xa}\\\upsilon_1=\frac{-5ax}{70-5ax}=\frac{-5ax}{-5(ax-14)}\\\boxed{\upsilon_1=\frac{ax}{ax-14} }$

$\upsilon_2=\frac{\partial f (x,a)}{\partial x}\cdot \frac{a}{f(x,a)}\\\upsilon_2=-5a\cdot \frac{a}{70-5xa}\\\upsilon_2=\frac{-5a^2}{70-5ax}=\frac{-5a^2}{-5(ax-14)}\\\boxed{\upsilon_2=\frac{a^2}{ax-14} }$

8 0

4 years ago