Answer:
No
Stepbystep explanation:
substitute 2 for 'x' and 5 for 'y'
2(2) + 5(5) = 0?
4 + 25 = 21, not 0
half person).
Answer:
40.62
Stepbystep explanation:
that's the answer I got
Pls help I will give brainliest
Answer:
x= 4/16 stan
x=0.26 dec
Stepbystep explanation:
⬇ Hello! The answer would be down below! :) ⬇
x = 0 will be the answer.
Stepbystep explanation:
So, we start with .
For step 1 we should subtract on both sides.
=
For step 2 we will multiply both sides by .
x x
= x = 0
16
12
8
20 16 12

4
48
12
16 2024
8
12
16
If the graph of the second equation in the system passes through (12, 20) and (4,12), which statement is true?
(a) For n = 6, CL = 90%,
The degrees of freedom: 5, Critical tvalue: 2.571
(b) For n = 21, CL = 98%,
The degrees of freedom: 20, Critical tvalue: 2.845
(c) For n = 29, CL = 95%,
The degrees of freedom: 28, Critical tvalue: 2.048
(d) For n = 12, CL = 99%,
The degrees of freedom: 11, Critical tvalue: 3.106
Use the concept of critical t value defined as:
A critical value is a number that is used in hypothesis testing to compare to a test statistic and evaluate whether or not the null hypothesis should be rejected. The null hypothesis cannot be rejected if the test statistic's value is less extreme than the crucial value.
(a) Given that,
n = 6 and a confidence level of 90%,
The degrees of freedom are,
n1 = 61
The degrees of freedom = 5.
To find the critical tvalue,
Look it up in the tdistribution table using a confidence level of 90% and a degree of freedom of 5.
From the table,
The critical tvalue is approximately 2.571.
(b) Given that,
n = 21 and a confidence level of 98%,
The degrees of freedom are,
n1 = 211
The degrees of freedom = 20.
By referring to the tdistribution table with a confidence level of 98% and degrees of freedom of 20,
The critical tvalue is approximately 2.845.
(c) Given that,
n = 29 and a confidence level of 95%,
The degrees of freedom are,
n1 = 291
The degrees of freedom = 28
Using the tdistribution table with a confidence level of 95% and degrees of freedom of 28,
The critical tvalue is approximately 2.048.
(d) Given that,
n = 12 and a confidence level of 99%,
The degrees of freedom are,
n1 = 121
The degrees of freedom = 11
By consulting the tdistribution table with a confidence level of 99% and degrees of freedom of 11,
The critical tvalue is approximately 3.106.
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To find the degrees of freedom and critical tvalue for each given sample size and confidence level, we can use the tdistribution and a ttable. The degrees of freedom (df) for each sample is equal to the sample size minus 1. The critical tvalue can be found using the ttable with the corresponding degrees of freedom and the confidence level.
To find the degrees of freedom and critical tvalue for each given sample size and confidence level, we can use the tdistribution and a ttable. The degrees of freedom (df) for each sample is equal to the sample size minus 1. For example, for (a) n = 6, df = 6  1 = 5. The critical tvalue can be found using the ttable with the corresponding degrees of freedom and the confidence level.
For (a) n = 6, CL = 90%, the critical tvalue is approximately 1.943.
For (b) n = 21, CL = 98%, the critical tvalue is approximately 2.861.
For (c) n = 29, CL = 95%, the critical tvalue is approximately 2.045.
For (d) n = 12, CL = 99%, the critical tvalue is approximately 3.106.
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The sequence 2, 8, 32, 128, . . . is geometric.
The sequence 5, 10, 15, 20, . . . is geometric.
The sequence 3, 18, 108, 648, . . . is geometric.
Answer:
1,2,3,5
Stepbystep explanation: