Answer: 1. rounds down to 1
2. rounds up to 10
3. rounds down to 7
4. not sure I say it rounds down to 4
5. rounds down to 2
6. rounds down to 6
7. rounds down to 5
8. rounds up to 4
The statement for geometric model using algebra tiles represents the factorization is the statement number 2.
<h3>What is algebra tile?</h3>
Algebra tiles used to represent the algebraic expression in the table form. The shape of algebra tile is square and rectangle, in which the variables represented.
The given polynomial equation in the problem is,
In the above polynomial, the highest power of variable is 2. Thus, it is a quadratic equation.The image of the algebra tile for the given polynomial attached below.
The statement which satisfy the polynomial and its algebra tile is,
An algebra tile configuration.
- 3 tiles are in the Factor 1 spot: 1 is labeled x and 2 are labeled negative.
- 3 tiles are in the Factor 2 spot: 1 is labeled x and 2 are labeled negative.
- 11 tiles are in the Product spot: 1 is labeled x squared, 4 are labeled negative x, and 6 are labeled.
Thus, the statement for geometric model using algebra tiles represents the factorization is the statement number 2.
Learn more about the algebra tile here;
brainly.com/question/4407619
Answer:
parallel = 4/3
perpendicular =-3/4
Step-by-step explanation:
Solve for y
-4x+3y=11
Add 4x to each side
3y = 4x+11
Divide by 3
y = 4/3 x +11/3
This is in the form y = mx+b where m is the slope
m =4/3
The parallel line has the same slope
parallel slope is 4/3
The perpendicular line has a negative reciprocal slope
m = -(1 /4/3) = - 3/4
<span>3√1000 = 3*31.623 = 94.87
We can also do this by doing </span><span>√9000 which is also 94.87</span>
Answer:
t=5.5080( to 3 d.p)
Step-by-step explanation:
From the data given,
n =20
Deviation= 34/20= 1.7
Standard deviation (sd)= 1.3803(√Deviation)
Standard Error = sd/√n
= 1.3803/V20 = 0.3086
Test statistic is:
t = deviation /SE
= 1.7/0.3086 = 5.5080
ndf = 20 - 1 = 19
alpha = 0.01
One Tailed - Right Side Test
From Table, critical value of t =2.5395
Since the calculated value of t = 5.5080 is greater than critical value of t = 2.5395, the difference is significant. Reject null hypothesis.
t score = 5.5080
ndf = 19
One Tail - Right side Test
By Technology, p - value = 0.000
Since p - value is less than alpha , reject null hypothesis.
Conclusion:
From the result obtained it can be concluded that ,the data support the claim that the mean rating assigned to the wine when the cost is described as $90 is greater than the mean rating assigned to the wine when the cost is described as $10.