How do I do the question

The graph of a proportional relationship passes through (12, 16)<br> and (1, y)<br> . Find y

k0ka [10]

Answer: The value of y is $\frac{4}{3}$ .

Explanation:

It is given that the graph of a proportional relationship passes through (12, 16)

and (1, y).

The graph of a proportional relationship means the x and y coordinates are in a proportion k. The equation of the graph is in the form of y=kx. Where k is the proportion factor.

It is given that the graph passing through (12,16).

$y=kx$

$16=k(12)$

$k=\frac{16}{12}$

$k =\frac{4}{3}$

So the equation of the line is,

$y=\frac{4}{3} x$

put x=1.

$y=\frac{4}{3} (1)$

$y=\frac{4}{3}$

Therefore, the value of y is $\frac{4}{3}$ .

8 0

3 years ago

Read 2 more answers

Last year, 31 2 acres of strawberry fields produced approximately 14,700 pounds of strawberries. This year the number of acres p

USPshnik [31]

So in 5 acres of land produces 21,000 pounds of strawberries.

Step-by-step explanation:

Step 1:

Last year, $3\frac{1}{2}$ acres i.e 3.5 acres produced 14,700 pounds of strawberries. So we need to determine how many strawberries were produced per acre.

To do this we divide the total number of strawberries produced by the number of acres it was produced in.

The number of strawberries produced per acre $= \frac{14,700}{3.5} = 4,200$ pounds.

Step 2:

To determine the number of strawberries in 5 acres of land, we multiply the number of strawberries per acre by the number of acres.

The number of strawberries produced in 5 acres $= 5(4,200) = 21,000$ pounds.

So in 5 acres of land produces 21,000 pounds of strawberries.

5 0

3 years ago

A student repeatedly measures the mass of an object using a mechanical balance and gets the following values: 560 g, 562 g, 556

MrRissso [65]

Answer: 2.76 g

Step-by-step explanation:

The formula to find the standard deviation:-

$\sigma=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{n}}$

The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.

Then, $\overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560$