Ask question

Ask question

All categories

3 years ago

12

An angle theta is in __ _ if its vertex is at the origin of a rectangular coordinate system and its initial side co

incides with the positive x-axis

1 answer:

romanna [79]3 years ago

6 0

We know that in standard position, the vertex of the angle should lie at the origin and its initial side is always fix along the positive x- axis. Please see the attached image.

We can see that the vertex of the angle is at the origin O and the initial side OB is along the positive x axis. This is the standard position of an angle.

Therefore, we have

An angle theta is in standard position if its vertex is at the origin of a rectangular coordinate system and its initial side coincides with the positive x-axis.

You might be interested in

WILL MARK BRAINLIST FOR ANSWER!!!!!!!!!!<br> |2 1/3 x+0.3|=3.5

Natalija [7]

THE ANSWER ISSSS

1.37142

6 0

3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

3 years ago

How many solutions does the following equation have?<br> -5(2+1) = –22 + 10

Igoryamba

Answer:

0 real # solution

Step-by-step explanation:

b^2-4ac<1 is equal to 0 real number solution and

-5(2+1)=-22+10

-5(2+1)=-12

which is less than 1

7 0

3 years ago

Divide 25.84 divide by 4

Nadusha1986 [10]

I believe it is 6.46

6 0

3 years ago

Read 2 more answers

In Bobby’s school, math grades for the year are calculated from assignments, tests and a final exam. Assignments count 30%, test

jarptica [38.1K]

Answer:

A) Bobby's overall grade is 70.4%

Step-by-step explanation:

To solve this problem we must calculate the percentages each grade represents from the overall grade. Therefore:

assignment = 85*30% = 85*30/100 = 25.5%

tests = 72*20% = 72*20/100 = 14.4%

exam = 61*50% = 61*50/100 = 30.5%

total = assignment + tests + exam

total = 25.5 + 14.4 + 30.5 = 70.4

Bobby's overall grade is 70.4%. Therefore the correct answer is A).

7 0

3 years ago