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2 years ago

Carolyn bought 9 feet of copper pipe for 54$ what was the cost of each foot of copper pipe?

Mathematics

2 answers:

Irina18 [472]2 years ago

8 0

Answer:

54÷9= 6

so its $6 a foot of copper

olya-2409 [2.1K]2 years ago

6 0

Ok so what you are going to do is divide 54 by 9 and you should get 6 now you have your answer $6.00

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Tom taxis charges a fixed rate $4 per ride plus $0.50 per mile Carla cabs does not charge a fixed rate but charges $1.00 per mil

RSB [31]

at 8 miles they are the same amount after that Carlas cab is more money than tims

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2 years ago

Read 2 more answers

Help me please<br> It about linear equations

marin [14]

Answer:

Hours: Elevation:

0 3000

2 2000

5 500

Step-by-step explanation:

Time is always placed on the x (horizontal) axis, and the dependent variable on the y. Just find the number given and find the corresponding value to go with it! Hope this helps :)

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3 years ago

Pani says she should get $3 discount on the price of each shirt and a $3 discount on the price of each pair of jeans. Write and

PilotLPTM [1.2K]

Answer:

(S - 3) + (X - 3) = B

Step-by-step explanation:

S = amount of shirts

X = amount of jeans

B = total amount she would pay

(S - 3) + (X - 3) = B

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3 years ago

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

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3 years ago

The graph of f passes through (-6,9) and is perpendicular to the line that has an x-intercept of 8 and a y-intercept of -24 the

Valentin [98]

Given:

The graph of f passes through (-6,9).

It is perpendicular to the line that has an x-intercept of 8 and a y-intercept of -24.

To find:

The equation of the function f.

Solution:

The equation of line on which graph of f is perpendicular, is

$\dfrac{x}{a}+\dfrac{y}{b}=1$

where, a and b are x and y intercepts respectively.

$\dfrac{x}{8}+\dfrac{y}{-24}=1$

Multiply both sides by 24.

$3x-y=24$

Slope intercept form is

$y=3x-24$

Slope of this line is 3 and y-intercept is -24.

Product of slopes of two perpendicular lines is -1.

Let the slope of f is m. So,

$m\times 3=-1$

$m=-\dfrac{1}{3}$

Slope of m is -1/3 and it passes through (-6,9). So, the equation of function f is

$y-y_1=m(x-x_1)$

$y-9=-\dfrac{1}{3}(x-(-6))$

$y-9=-\dfrac{1}{3}(x+6)$

$y-9=-\dfrac{1}{3}x-2$

$y=-\dfrac{1}{3}x-2+9$

$y=-\dfrac{1}{3}x+7$

Put y=f(x).

$f(x)=-\dfrac{1}{3}x+7$

Therefore, the required function is $f(x)=-\dfrac{1}{3}x+7$ .

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2 years ago