the brunch buffet cost the same for each adult. for the family of 8 the total cost was 272 how much did it cost per person

In circle N with m<br> See diagram below

Alex777 [14]

Answer:

$113.10$

Step-by-step explanation:

The area of a sector with measure $\theta$ and radius $r$ is given by $A_{sec}=r^2\pi\cdot \frac{\theta}{360^{\circ}}$ .

What we're given:

$r$ of 12
$\theta$ of $90^{\circ}$

Substituting given values, we get:

$A_{sec}=12^2\pi\cdot \frac{90}{360},\\\\A_{sec}=144\pi\cdot \frac{1}{4},\\\\A_{sec}\approx \boxed{113.10}$

8 0

3 years ago

Guys i need help on math plz

olga_2 [115]

Answer:

look at the picture for the answer

7 0

2 years ago

What is the value of x?? please help

DaniilM [7]

Answer:

x=64

Step-by-step explanation:

Both the semicircle that says 180, and the right angle sign in the image serve to support the idea that all parts of the figure equal 90

Therefore, it is safe to say that you need to subtract 26 from 90 in order to solve for x.

90-26 = 64

8 0

3 years ago

Read 2 more answers

Each point on the coordinate plane has an address called

RideAnS [48]

I do not understand what to do , more explicitly sorry :(

8 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