Answer:
x=-1
Step-by-step explanation:
x+2-1=0
-1+2=1
1-1=0
y=0
The answer is 44
To find this answer, we replace x with 8 and use PEMDAS (order of operations) to simplify
f(x) = 6*x - 4
f(8) = 6*8 - 4
f(8) = 48 - 4
f(8) = 44
Answer:
RADIUS
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PROBLEM
Mary’s bicycle wheel has a circumference of 226.08 cm². What is its radius?
SOLUTION
We can solve this problem using the circumference formula in which π stands for ( 3.14 ), C stands for circumference itself and r stands for radius.
\bold{Formula \: || \: C = 2πr}Formula∣∣C=2πr
\tt{226.08 = 2(3.14) r}226.08=2(3.14)r
'Now to find the radius,Substitute 226.08 for c which is circumference in the formula.
\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{C = 2πr} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{226.08 = 2(3.14)\red{r}} \\ \\ \: \: \: \: \: \: \: \: \large \tt{ \frac{226.08}{6.28} = \cancel\frac{6.28 \red{r}}{6.28} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt\green{C = 36}}\end{gathered}
C=2πr
226.08=2(3.14)r
6.28
226.08
=
6.28
6.28r
C=36
To check:
\begin{gathered} \small\begin{array}{|c|}\hline \bold{circumference }\\ \\ \tt{C = 2πr} \\ \tt{C = 2(3.14) (36\:cm) } \\ \tt{C = 2(113.04\:cm) } \\ \underline{\tt \green{C = 226.08\:cm }} \\ \hline \end{array} \end{gathered}
circumference
C=2πr
C=2(3.14)(36cm)
C=2(113.04cm)
C=226.08cm
FINAL ANSWER
If Mary's Bicycle has a circumference of 226.08 cm then the radius is 36.
\boxed{ \tt \red{r = 36}}
r=36
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#CarryOnLearning
I am going to do number 2 so when you subtract decimals you have to add two zeros like this 10.00 or 0.25 so now when you subtract you get $9.75
Answer:
Since the calculated value of F = 1.4397 is less than the critical value of
F (9,9)= 2.4403 we conclude that the first instructor's variance is smaller and reject H0.
Step-by-step explanation:
1)Formulate the hypothesis that first variance is equal or greater than the second variance
H0: σ₁²≥σ₂² against the claim that the first instructor's variance is smaller
Ha: σ₁²< σ₂²
2) Test Statistic F= s₂²/s₁²
F= 84.8/ 58.9= 1.4397
3)Degrees of Freedom = n1-1= 10-1= 9 and n2 = 10-1= 9
4)Critical value at 10 % significance level= F(9,9)= 2.4403
5)Since the calculated value of F = 1.4397 is less than the critical value of
F (9,9)= 2.4403 we conclude that the first instructor's variance is smaller and reject H0.
