The temperature was -3c last night. it is now -4c. what was the change in temperature

The equation y=0.5x+40 represents the monthly cost y in dollars of Lesley's cell phone, where x is the of talk minutes over 750

Firdavs [7]

You already have the equation given which is <span>y=0.5x+40. To graph, you just have to replace random values of x to determine the corresponding values of y. Plot these points and connect them. The graph is shown in the attached picture. As you can observe, the range starts from y=40. This is because the y-intercept is 40. So, you don't have to show the y-values below because it would just minimize your linear graph.</span>

7 0

3 years ago

LCM OF 24,39,60 and 150 is<br>

Salsk061 [2.6K]

Answer:

The LCM of 24,39,60, and 150 is 7,800.

Step-by-step explanation:

I use prime factorization and then I mulitply all of the double numbers.

8 0

3 years ago

he data represents the daily rainfall (in inches) for one month. Construct a frequency distribution beginning with a lower clas

Licemer1 [7]

Answer:

It is not normally distributed as it has it main concentration in only one side.

Step-by-step explanation:

So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).

So, let us begin the groupings into their different classes, shall we?

Data given:

0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.

(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.

(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31

(3). 0.4 - 0.6 : there are 2 values that falls into this category.

(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.

Class interval frequency.

0.00 - 0.20. 15.

0.20 - 0.40. 2.

0.4 - 0.6. 2.

4 0

3 years ago

I need help I can’t figure it out

dusya [7]

Answer:

$\large\boxed{\dfrac{41p}{70q^4(r-9)^2}}$

Step-by-step explanation:

$\dfrac{2p^4}{5q^5(r-9)^3}\cdot\dfrac{41q(r-9)}{28p^3}\\\\=\dfrac{2\!\!\!\!\diagup^1\cdot41}{5\cdot28\!\!\!\!\!\diagup_{14}}\cdot\dfrac{q\!\!\!\!\diagup}{q^{5\!\!\!\!\diagup^4}}\cdot\dfrac{(r\!\!\!\!\!\!{--}-9)\!\!\!\!\!\!{--}}{(r-9)^{3\!\!\!\!\diagup^2}}\cdot\dfrac{p^{4\!\!\!\!\diagup}}{p^3\!\!\!\!\!\!\diagup}\\\\=\dfrac{41}{70}\cdot\dfrac{1}{q^4}\cdot\dfrac{1}{(r-9)^2}\cdot\dfrac{p}{1}\\\\=\dfrac{41p}{70q^4(r-9)^2}$

7 0

3 years ago

Rewrite the system of linear equations as a matrix equation AX = B.

iren2701 [21]

Answer:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]$

And the vector of vriables (X) is:

$\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]$

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

$\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

In conclusion, AX=B is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

7 0

3 years ago