Determine the domain and range of:

If tan x =3/4, and 0

jasenka [17]

Answer: x=0.6435011

Step-by-step explanation:

Take the inverse tangent of both sides of the equation to extract

x

from inside the tangent.

x

=

arctan

(

3

4

)

Evaluate

arctan

(

3

4

)

.

x

=

0.6435011

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from

π

to find the solution in the fourth quadrant.

x

=

(

3.14159265

)

+

0.6435011

Simplify the expression to find the second solution.

Tap for more steps...

x

=

3.78509376

Find the period.

Tap for more steps...

π

The period of the

tan

(

x

)

function is

π

so values will repeat every

π

radians in both directions.

x

=

0.6435011

+

π

n

,

3.78509376

+

π

n

, for any integer

n

Consolidate the answers.

x

=

0.6435011

+

π

n

, for any integer

n

6 0

3 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago

Please help, photo attached

AlexFokin [52]

Answer:

Step-by-step explanation:

B = -2

8 0

3 years ago

The rectangle shown has a perimeter of 7272 cm and the given area. its length is 66 more than four timesfour times its width. wr

Vadim26 [7]

<span>Width = 6 Length = 30 We know the perimeter of a rectangle is simply twice the sum of it's length and width. So we have the expression: 72 = 2*(L + W) And since we also know for this rectangle that it's length is 6 more than 4 times it's width, we have this equation as well: L = 6 + 4*W So let's determine what the dimensions are. Since we have a nice equation that expresses length in terms of width, let's substitute that equation into the equation we have for the perimeter and solve. So: 72 = 2*(L + W) 72 = 2*(6 + 4*W + W) 72 = 2*(6 + 5*W) 72 = 12 + 10*W 60 = 10*W 6 = W So we now know that the width is 6. And since we have an expression telling us the length when given the width, we can easily determine the length. So: L = 6 + 4*W L = 6 + 4*6 L = 6 + 24 L = 30 And now we know the length as well.</span>

5 0

3 years ago

PLEASE HELP! AS SOON AS POSSIBLE I HAVE TO BE DONE IN 10 MINS !

jolli1 [7]

Answer:

D

Step-by-step explanation:

6 0

3 years ago

Determine the domain and range of:​

Determine the domain and range of: