Answer:
The correct answer is option (C)-0.245 = 2.160(0.205)
Step-by-step explanation:
Solution
Given that:
The slope = - 0.245
The size sample = n = 15
The standard error = 0.205
The confidence level = 95
The Significance level= α = (100- 95)% = 0.05
Now,
The freedom of degree = n-2 = 15 -2= 13
Thus,
the critical value = t* = 2.16
By applying Excel = [TINV (0.05, 13)]
The Margin of error is = t* (standard error)
=2.16 *0.205
= 0.4428
Answer:

General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS<u>
</u>
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
- Coordinates (x, y)
- Slope Formula:

Step-by-step explanation:
<u>Step 1: Define</u>
Point (2, 5)
Point (-3, 7)
<u>Step 2: Identify</u>
x₁ = 2, y₁ = 5
x₂ = -3, y₂ = 7
<u>Step 3: Find slope </u><em><u>m</u></em>
Simply plug in the 2 coordinates into the slope formula to find slope<em> m</em>
- Substitute in points [Slope Formula]:

- [Slope] [Fraction] Subtract:

- [Slope] [Fraction] Rewrite:

98cm^2 - bigger triangles are 28 each, bottom one is 14. (28 x 3) + 14=98
Answer:

Step-by-step explanation:
The formula for the accrued amount from compound interest is

1. Amount in account on 1 Jan 2015
(a) Data:
a = £23 517.60
r = 2.5 %
n = 1
t = 1 yr
(b) Calculations:
r = 0.025

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.
She must have had £23 944 in her account on 1 Jan 2015.
(2) Amount originally invested
(a) Data
A = £23 944.00

3. Summary
1 Jan 2014 P = £23 360.00
1 Jan 2015 A = 23 944.00
Withdrawal = <u> -1 000.00
</u>
P = 22 944.00
1 Jan 2016 A = £23 517.60