Step-by-step explanation:
Next term is - 51
Use formula Tn=a+(n-1)d
a=the first term of the sequence which is - 15
d=the difference between the terms which is - 9
Substitute
Tn=-15+(n-1)(-9)
Tn=-15+9-9n
Tn=-9n-6
From this you will be able to generate all the terms in the pattern
If they say determine the 50th term just substitute 50 by n
Tn=-9(50)-6
Tn=-450-6
Tn=-456
A Credit union is a group of people who agree to save their money together and to make loans to each other at a relatively low rate of interest. <span />
p.s. Excuse the scribble.
Step-by-step explanation:
This is a piece-wise function. The three intervals we need to worry about are [0, 1), [1,2], and [2,4].
Separate the functions into their pieces and draw out the individual graphs. Place them together onto the graph within their respective intervals.
Answer:
180 Square Feet
Step-by-step explanation:
For Rectangle:
l = 9 ft
b = 20 ft
Area of Rectangle

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)