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2 years ago

F (x) = √x + 9 what is the domain?

2 answers:

7 0

Hopes this helps:

Answer: x ≥ 0

Anything inside a square root must not be negative. The domain has to be x ≥ 0.

And if you have to graph it it has to start at 0 and go up to 9 and curve over.

Gala2k [10]2 years ago

4 0

Answer:

Step-by-step explanation:

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Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering

Alla [95]

Answer:

a) The probability that Jodi scores 78% or lower on a 100-question test is 4%.

b) The probability that Jodi scores 78% or lower on a 250-question test is 0.023%.

Step-by-step explanation:

a) To approximate this distribution we have to calculate the mean and the standard distribution.

The mean is the proportion p=0.85.

The standard deviation can be calculates as:

$\sigma=\sqrt{\frac{p(1-p)}{n} }= \sqrt{\frac{0.85*(1-0.85)}{100} }=0.04$

To calculate the probability that Jodi scores 78% or less on a 100-question test, we first calculate the z-value:

$z=\frac{p-p_0}{\sigma} =\frac{0.78-0.85}{0.04} =-1.75$

The probability for this value of z is

$P(x$

The probability that Jodi scores 78% or lower on a 100-question test is 4%.

b) In this case, the number of questions is 250, so the standard deviation needs to be calculated again:

$\sigma=\sqrt{\frac{p(1-p)}{n} }= \sqrt{\frac{0.85*(1-0.85)}{250} }=0.02$

To calculate the probability that Jodi scores 78% or less on a 250-question test, we first calculate the z-value:

$z=\frac{p-p_0}{\sigma} =\frac{0.78-0.85}{0.02} =-3.5$

The probability for this value of z is

$P(x$

The probability that Jodi scores 78% or lower on a 250-question test is 0.023%.

6 0

3 years ago

Find the point-slope form of the equation of the line that passes through the given point and has the indicated slope. Point: (-

zzz [600]

Answer:

slope : 3

y + 3 = 3(x + 5)

y + 3 = 3x + 15

y = 3x + 12

5 0

3 years ago

{(-3,9), (-2, 4), (0, 0), (1, 1)}<br> Find the inverse

Elena-2011 [213]

Answer:

Step-by-step explanation:

to find the inverse, just switch x and y

so if u have these points : (-3,9) , (-2,4), (0,0), (1,1)

the inverse would be : (9,-3) , (4,-2), (0,0), (1,1)

5 0

3 years ago

An investigator wants to assess whether the mean m = the weight of passengers flying on small planes exceeds the FAA guideline o

stira [4]

Answer:

$H_{0}: \mu= 185$ and $H_{a}: \mu > 185$

Step-by-step explanation:

The null hypothesis $H_{0}$ states that a population parameter (such as the mean, the standard deviation, and so on) is equal to a hypothesized value. We can write the null hypothesis in the form $H_{0}: parameter = value$

In this context, the investigator's null hypothesis should be that the average total weight is no different than the reported value by the FAA. We can write it in this form $H_{0}: \mu= 185$ .

The alternative hypothesis $H_{a}$ states that a population parameter is smaller, greater, or different than the hypothesized value in the null hypothesis. We can write the alternative hypothesis in one of three forms

$H_{a}: parameter > value\\H_{a}: parameter < value\\H_{a}: parameter \neq value$

The investigator wants to know if the average weight of passengers flying on small planes exceeds the FAA guideline of the average total weight of 185 pounds. He should use $H_{a}: \mu > 185$ as his alternative hypothesis.

7 0

3 years ago

Find the solution of the differential equation dy/dt = ky, k a constant, that satisfies the given conditions. y(0) = 50, y(5) =

irga5000 [103]

Answer: The required solution is $y=50e^{0.1386t}.$

Step-by-step explanation:

We are given to solve the following differential equation :

$\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.

From equation (i), we have

$\dfrac{dy}{y}=kdt.$

Integrating both sides, we get

$\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]$

Also, the conditions are

$y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50$

and

$y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.$

Thus, the required solution is $y=50e^{0.1386t}.$

8 0

3 years ago

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