Answer:
Step-by-step explanation:
Given that
Group Group One Group Two
Mean 26.00 23.00
SD 2.00 4.00
SEM 0.40 1.21
N 25 11
where group I represents female servers and group II male servers.
We have to calculate confidence interval for 90% for difference in means
The mean of Group One minus Group Two equals 3.00
df = 34
standard error of difference = 0.993
t critical = 2.034 for 90% df 34
Hence confid. interval at 90%
=Mean diff ±2.034 * std error of diff
= (0.98, 5.02)
(x - 1)(x - 2)(x + 2)
note that the sum of the coefficients 1 - 1 - 4 + 4 = 0
thus x = 1 is a root and (x - 1 ) is a factor
dividing x³ - x² - 4x + 4 by (x - 1)
x³ - x² - 4x + 4 = (x - 1)(x² - 4 ) (note (x² - 4 ) is a difference of squares )
x³ - x² - 4x + 4 = (x - 1)(x - 2)(x + 2)
(x - 1)(x - 2)(x + 2 ) =0
x - 1 = 0 ⇒ x = 1
x - 2 = 0 ⇒ x = 2
x + 2 = 0 ⇒ x = - 2
solutions are x = 1 or x = ± 2
<h3>
Answer: Negative</h3>
======================================================
Explanation:
3/7 = 0.42857 approximately
Pick a number between that value and 5, not including either endpoint. Let's say we pick x = 2
Plug x = 2 into the f(x) function
f(x) = (x - 5)(2x + 7)(7x-3)
f(2) = (2 - 5)(2*2 + 7)(7*2-3)
f(2) = (2 - 5)(4 + 7)(14-3)
f(2) = (-3)(11)(11)
f(2) = -363
The actual result doesn't matter. All we're after is whether the result is positive or negative. We see the result is negative. This means f(x) is negative when 3/7 < x < 5. The f(x) curve is below the x axis on this interval.
<u>Answer:</u><u> </u><u>3</u><u>4</u><u> </u><u>m</u><u>o</u><u>r</u><u>e</u><u> </u><u>p</u><u>o</u><u>u</u><u>c</u><u>h</u><u>e</u><u>s</u>
7 boxes - fish food; 1 box contained 6 pouches
Total amount of fish food is 7 × 6 = 42 pouches of fish food
2 packets - cat food; 1 packet contained 4 pouches
Total amount of cat food is 2 × 4 = 8 pouches
How many more pouches of fish food than cat food did Shelly buy?
42 pouches - 8 pouches = 34 pouches
Therefore there was 34 more pouches of fish food than cat food
