Answer:
a) Statement is true
b) Statement incorrect. Company goal is to attend customer at 190 second at the most. So test should be one tail test (right)
c) We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error
Step-by-step explanation:
a) as the significance level is = 0,02 that means α = 0,02 (the chance of error type I ) and the β (chance of type error II is
1 - 0.02 = 0,98
b) The company establishe in 190 seconds time for customer at its ticket counter (if this time is smaller is excellent ) company is concerned about bigger time because that could be an issue for customers. Therefore the test should be a one test-tail to the right
c) test statistic
Hypothesis test should be:
null hypothesis H₀ = 190
alternative hypothesis H₀ > 190
t(s) = ( μ - μ₀ ) / s/√n ⇒ t(s) = ( 202 - 190 )/(28/√100 )
t(s) = 12*10/28
t(s) = 4.286
That value is far away of any of the values found for 99 degree of fredom and between α ( 0,025 and 0,01 ). We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error
If we look table t-student we will find that for 99 degree of freedom and α = 0.02.
Answer: 12,150 m²
<u>Step-by-step explanation:</u>
The ratio is 3:2 which means
Length (L) = 3x
Width (w) = 2x
Perimeter (P) = 2L + 2w
450 = 2(3x) + 2(2x)
450 = 6x + 4x
450 = 10x
45 = x
L = 3x w = 2x
= 3(45) = 2(45)
= 135 = 90
Area (A) = L × w
= 135 × 90
= 12,150
Gretchen ran 48 meters in nine seconds, x+time (9) and y= meters (48)
i hope this helped :)
Answer:
see explanation
Step-by-step explanation:
(a)
Given
2k - 6k² + 4k³ ← factor out 2k from each term
= 2k(1 - 3k + 2k²)
To factor the quadratic
Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)
The factors are - 1 and - 2
Use these factors to split the k- term
1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )
1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term
= (1 - k)(1 - 2k)
1 - 3k + 2k² = (1 - k)(1 - 2k) and
2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)
(b)
Given
2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )
= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term
= (x - 2y)(2a + 3b)
