No I don’t think so (&/&/ need extra characters)
< o and < m are congruent because opposite angles are congruent
3t - 15 = 2t + 10
3t - 2t = 10 + 15
t = 25
3t - 15 = 3(25) - 15 = 75 - 15 = 60...< m = 60
< m and < o are consecutive angles and they are supplementary....so they add up to 180
< m + < o = 180
60 + < o = 180
< o = 180 - 60
< o = 120
3s = 120
s = 120/3
s = 40
As per the isosceles triangle theorem, if the two base angles are congruent then the legs are also congruent, so we must set the two legs equal to each other:
2z - 15 = 9
Add 15 to both sides:
15 - 15 = 0
15 + 9 = 24
Divide 2 from each side:
2z = 24
2z/2 = z
24/2 = 12
z = 12
Hence, the answer would be D: z = 12
Answer:
<h2>3) -82.</h2><h2>4)-70.</h2><h2>5) -15.</h2><h2>6)-14.</h2><h2>7)5</h2><h2>8)15</h2><h2>9)16</h2><h2>10 )16</h2><h2>11)64</h2><h2>12)21.</h2><h2>13)1500 in the middle (not above not below)</h2><h2>14)450 m</h2>
Answer:
a. We reject the null hypothesis at the significance level of 0.05
b. The p-value is zero for practical applications
c. (-0.0225, -0.0375)
Step-by-step explanation:
Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.
Then we have 
, 
, 
and 
, 
, 
. The pooled estimate is given by
a. We want to test 
vs 
(two-tailed alternative).
The test statistic is 
and the observed value is 
. T has a Student's t distribution with 20 + 25 - 2 = 43 df.
The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value 
falls inside RR, we reject the null hypothesis at the significance level of 0.05
b. The p-value for this test is given by 
0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)
, i.e.,
where 
is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
, i.e.,
(-0.0225, -0.0375)
