A triangle is 3cm wider than it is tall. The area is 44cm^2. Find the width(length of the base).

Can you help me on 34 and 35 please

dusya [7]

-- #34 is choice-C .

-- See the picture attached to this answer, for both solutions.

6 0

2 years ago

2. What conjecture can you make about the eighteenth term in the pattern A, B, A, C, A, B, A, C?

slamgirl [31]

Answer:

The eighteenth term is B

Step-by-step explanation:

I could just write the pattern A B A C out until you get to the eighteenth term

A(1) B(2) A(3) C(4) A(5) B(6) A(7) C(8) A(9) B(10) A(11) C(12) A(13) B(14) A(15) C(16) A(17) B(18)

4 0

2 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:

$\bar x = 70.2903$