Answer:
- addition property of equality
- integers are closed to addition
- identity element
- multiplication property of equality
- commutative property of multiplication; reals are closed to multiplication; identity element
Step-by-step explanation:
<u>Given</u>:
c/2 -5 = 7
Step 1: c/2 -5 +5 = 7 +5
Step 2: c/2 +0 = 12
Step 3: c/2 = 12
Step 4: 2(c/2) = 12(2)
Step 5: c = 24
<u>Find</u>:
The property that justifies each step of the solution.
<u>Solution</u>:
Step 1: addition property of equality (lets you add the same to both sides)
Step 2: integers are closed to addition
Step 3: identity property of addition (adding 0 changes nothing)
Step 4: multiplication property of equality
Step 5: closure of real numbers to multiplication; identity property of multiplication
_____
It is hard to say what "property" you want to claim when you simplify an arithmetic expression. Above, we have used the property that the sets of integers and real numbers are closed to addition and multiplication. That is, adding or multiplying real numbers gives a real number.
In Step 5, we can rearrange 2(c/2) to c(2/2) using the commutative property of multiplication. 2/2=1, and c×1 = c. The latter is due to the identity element for multiplication: multiplying by 1 changes nothing.
Apart from the arithmetic, the other properties used are properties of equality. Those let you perform any operation on an equation, as long as you do it to both sides of the equation. The operations we have performed in this fashion are adding 5 and multiplying by 2.
Hello IdontKnowHowToMath,
first, converting R percent to r a decimal
r = R/100 = 3%/100 = 0.03 per year.
Solving our equation:
A = 6000(1 + (0.03 × 4)) = 6720
A = $6,720.00
The
total amount accrued, principal plus interest, from simple interest on a
principal of $6,000.00 at a rate of 3% per year for 4 years is
$6,720.00.
Answer:
-7k + 14
Step-by-step explanation:
(k² - 7k + 2) - (k² - 12)
k² - 7k + 2 - k² + 12
k² - k² - 7k + 2 + 12
-7k + 14
Answer:
z= 0.278
Step-by-step explanation:
Given data
n1= 60 ; n2 = 100
mean 1= x1`= 10.4; mean 2= x2`= 9.7
standard deviation 1= s1= 2.7 pounds ; standard deviation 2= s2 = 1.9 lb
We formulate our null and alternate hypothesis as
H0 = x`1- x`2 = 0 and H1 = x`1- x`2 ≠ 0 ( two sided)
We set level of significance α= 0.05
the test statistic to be used under H0 is
z = x1`- x2`/ √ s₁²/n₁ + s₂²/n₂
the critical region is z > ± 1.96
Computations
z= 10.4- 9.7/ √(2.7)²/60+( 1.9)²/ 100
z= 10.4- 9.7/ √ 7.29/60 + 3.61/100
z= 0.7/√ 0.1215+ 0.0361
z=0.7 /√0.1576
z= 0.7 (0.396988)
z= 0.2778= 0.278
Since the calculated value of z does not fall in the critical region so we accept the null hypothesis H0 = x`1- x`2 = 0 at 5 % significance level. In other words we conclude that the difference between mean scores is insignificant or merely due to chance.
