What is the name of a polygon with five sides?

What where William total fixed variable expenses the first year of college

VARVARA [1.3K]

Since the details are not available, we have provided an overview of William's total fixed and variable expenses for the first year of college.

<h3>What are college costs?</h3>

There are five categories of college costs.

These college costs include tuition, room and board, books and supplies, personal expenses, and transportation.

Some of the college costs like tuition and room and board may be relatively fixed.

This leaves expenses like books and suppliers, personal expenses, and transportation relatively variable.

Thus, William needs to control his variable college costs so that they do not exceed the student loan.

Learn more about college costs at brainly.com/question/11650418

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8 0

2 years ago

Express the complex number in trigonometric form.<br><br> -6 + 6\sqrt(3) i

horrorfan [7]

Answer:

Step-by-step explanation:

The rectangular form of a complex number is expressed as z = x+iy

where the modulus |r| = $\sqrt{x^{2}+y^{2}$ and the argument $\theta = tan^{-1}\frac{y}{x}$

In polar form, x = $rcos\theta \ and\ y = rsin\theta$

$z = rcos\theta+i(rsin\theta)\\z = r(cos\theta+isin\theta)$

Given the complex number, $z = -6+6\sqrt{3} i$ . To express in trigonometric form, we need to get the modulus and argument of the complex number.

$r = \sqrt{(-6)^{2}+(6\sqrt{3} )^{2}}\\r = \sqrt{36+(36*3)} \\r = \sqrt{144}\\ r = 12$

For the argument;

$\theta = tan^{-1} \frac{6\sqrt{3} }{-6} \\\theta = tan^{-1}-\sqrt{3} \\\theta = -60^{0}$

Since tan is negative in the 2nd and 4th quadrant, in the 2nd quadrant,

$\theta =180-60\\\theta = 120^{0}$

z = 12(cos120°+isin120°)

This gives the required expression.

4 0

4 years ago

A museum employee surveys a random sample of 350 visitors to the museum. Of those visitors, 266 stopped at the gift shop. Based

rjkz [21]

Answer: 1748.

Step-by-step explanation: This is because 266 is 76% of 350, so 76% of 2300 is 1748.

Hope that helped you out!

7 0

3 years ago

Hermes earns $6 an hour for babysitting. He wants to earn at least $168 for a new video game system. Determine the number of hou

solong [7]

Answer:

28 hours

Step-by-step explanation:

6x28=168

5 0

3 years ago

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

4 0

3 years ago