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3 years ago

How do you calculate correctly with real world measurements

1 answer:

6 0

Based on the experience of the responder, to correctly calculate measurements in real-world. Firstly is to avoid errors as much as possible. Errors are what makes your measurement invalid and unreliable. There are two types of error which is called the systematic error and the random error. Each error has different sources. Words that were mentioned –invalid and unreliable are very important key aspects to determine that your measure is truly accurate and consistent. Some would recommend using the mean method, doing three trials in measuring and getting their mean, in response to this problem.
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Someone please help me with this algebra problem

nevsk [136]

Answer:

$110

Step-by-step explanation:

We are looking for the decrease in money per desk purchased. Because the x-axis is the number of desks purchased, what we have to find is the (opposite) of the slope. The formula to represent slope can be written as:

$\frac{y_{2} -y_{1} }{x_{2}- x_{1} }$

We can substitute the points given, (2, 480) and (5, 150) and simplify:

$\frac{150-480}{5-2} =\frac{-330}{3} = -110$

We would multiply by -1 because we are asking for the decrease, and negative increase is decrease. -1 * -110, gives us $110

4 0

3 years ago

-9(x - 3) (Distributive property)

otez555 [7]

Answer:

-9x+27

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

If a/2b =3/5 and 4b/3c=1/7, then a/c= I will mark you brainlist whoever get this correct

ioda

Answer:

a/c=2/7 u are welcome thanks for ur help

7 0

2 years ago

a circular oil slick is expanding at a rate of 2m^2/h. Find the rate at which it's diameter is expanding when it's radius is 1.5

aleksandrvk [35]

Answer: $\frac{4\pi}{3} \text{ meter per hour}$

Step-by-step explanation:

The circular oil slick is expanding at a rate of $2 m^2/h$

Let A be the area of the circular oil slick,

So, the changes in A with respect to time (t),

$\frac{dA}{dt} = 2$

$\frac{d(\pi r^2)}{dt} = 2$

$2\pi r\frac{dr}{dt} = 2$

$\frac{dr}{dt} = \frac{1}{\pi r}$

Also, the change in diameter with respect to time(t),

$\frac{d}{dt} (2 r) = 2 \frac{dr}{dt}$

$\frac{d}{dt} (2 r) = 2 \times \frac{1}{\pi r}$

$\frac{d}{dt} (2 r) = \frac{2}{\pi r}$

For r = 1.5 m,

$\frac{d}{dt} ( 2 r)]_{r=1.5} = \frac{2}{\pi \times 1.5}=\frac{20}{\pi \times 15}=\frac{4\pi }{3}\text{ meter per hour}$

7 0

2 years ago

A total of 679 tickets were sold for the school play. they were either adult tickets or student tickets. there were 71 fewer stu

guapka [62]

For example a stands for the number of adult tickets, and s stands for the number of student tickets.

Write a system of equation based on the problem
An equation for "A total of 679 tickets were sold for the school play"
∴ a + s = 679 (first equation)
An equation for "there were 71 fewer student tickets sold than adult tickets"
∴ s = a - 71 (second equation)

Find the solution of the system
Subtitute s with a - 71 in the first equation to find the value of a
a + s = 679
a + a - 71 = 679
2a - 71 = 679
2a = 679 + 71
2a = 750
a = 750/2
a = 375
The number of adult tickets sold is 375

3 0

2 years ago