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1 year ago

If sin theta < 0 and tan theta > 0 then

1 answer:

7 0

If sin theta < 0 and tan theta > 0 as described in the task content, then, cos theta < 0 and 180<theta <270.

<h3>What is correct about angle theta as described?</h3>

From the task content, it was described that sin theta < 0 and tan theta > 0.

Upon recall of angle quadrants, it follows that only the 3rd quadrant angles have positive tan values and negative sin values. Hence, it follows that the angle theta is in the 3rd quadrant and also has cos theta < 0.

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Write $3.9 out of $60 as percentage

Norma-Jean [14]

Answer:

Step-by-step explanation:

$\frac{3.9}{60}*100=\frac{13}{2}$

= 6.5%

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1 year ago

Given the quadratic function g(x)=x^2, find g(3x-2).

maxonik [38]

Answer:

$g(3x-2) = (3x-2)^{2}$

Step-by-step explanation:

$g(x)=x^{2}$

To do this, you must substitute $3x-2$ for x in the equation g(x)

$g(3x-2) = (3x-2)^{2}$

Simplified this would be

$9x^{2} -12x+4$

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2 years ago

HELP QUICK LIKE RIGHT NOW!please

pogonyaev

Answer:

1/2 = 1/2

Step-by-step explanation:

5/10 is equal to 1/2 and 1/2 cant get smaller making you answer 1/2 = 1/2

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2 years ago

Read 2 more answers

What is the difference between a rhombus and a HURRY 5 MINUTES LEFT!!

Alik [6]

Answer:

A. rhombus is a parallelogram.

5 0

2 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

1 year ago