Solve for y when x = 5. y = 3x + 4

The vertices of the quadrilateral JKLM are J(-2,0), K(-1,2), L(1,3), and M(0,1). Can you use the slopes

qaws [65]

Answer:

A. Yes

B. Yes

C. No

Here's what it would look like:

6 0

3 years ago

Find the value of x and y

const2013 [10]

is the same thing here as the other triangles, the triangle below is an equilateral, the one above is an isosceles.

the vertex of the isosceles is 40°, so the twins angle at the base are (180 - 40)/2 = 70° = x each.

since the sides in the equilateral are all the same length, then 2y + 10 = x - y, but we already know what "x" is, so

$\bf 2y+10=x-y\implies 2y+10=\stackrel{x}{70}-y\implies 3y=60 \\\\\\ y=\cfrac{60}{3}\implies y=20$

6 0

3 years ago

HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE!

kiruha [24]

Answer:

b

i did this before xD

6 0

3 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$