Ask question

Ask question

All categories

3 years ago

5

A surf shop charges $6 to rent a wetsuit and $12 per hour to rent a surfboard. Write a function to model the cost, c, to rent a

wetsuit and surfboard for h hours. Enter the correct function in the box.

1 answer:

Over [174]3 years ago

8 0

C = 6 + $12^{x}$

You might be interested in

What can you say about the y-help urgent!!!

weeeeeb [17]

The correct answer is ( A )

let me know if you need explanation.

3 0

3 years ago

Will an exponential function always, never, or sometimes exceed a polynomial function?

patriot [66]

At some point, the exponential function always exceeds a polynomial function.

3 0

3 years ago

Read 2 more answers

A circle has a radius of 2.5 cm. What is the diameter of this circle in cm? Record your answer in Blank 1. What is the circumfer

Eva8 [605]

Answer:

The diameter is 5 cm and the circumference is diameter x pi

Step-by-step explanation:

3 0

3 years ago

Convert -3pi/10 to degrees

ipn [44]

we know there are 180° in π radians, how many degrees then in -3π/10 radians?

$\bf \begin{array}{ccll} degrees&radians\\ \cline{1-2} 180&\pi \\\\ x&-\frac{3\pi }{10} \end{array}\implies \cfrac{180}{x}=\cfrac{\pi }{~~-\frac{3\pi }{10}~~}\implies \cfrac{180}{x}=\cfrac{\frac{\pi}{1} }{~~-\frac{3\pi }{10}~~} \\\\\\ \cfrac{180}{x}=\cfrac{\pi }{1}\cdot \cfrac{10}{-3\pi }\implies \cfrac{180}{x}=-\cfrac{10}{3}\implies 540=-10x\implies \cfrac{540}{-10}=x \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill -54=x~\hfill$

5 0

3 years ago

Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow

prisoha [69]

Answer:

The answers is " Option B".

Step-by-step explanation:

$CI=\hat{Y}\pm t_{Critical}\times S_{e}$

Where,

$\hat{Y}=$ predicted value of lead content when traffic flow is 15.

$\to df=n-1=8-1=7$

$95\% \ CI\ is\ (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\$

$=\frac{1059.8}{2}\\\\=529.9$

Calculating thet-critical value $t_{ \{\frac{\alpha}{2},\ df \}}=-2.365$

The lower predicted value $=529.9-2.365(Se)$

$463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076$

When $99\%$ of CI use as the expected lead content: $\to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)$

8 0

3 years ago