Answer:
it will take 90 minutes
Step-by-step explanation:
10 - 6 = 9
9 x 10 = 90
Answer:
y = 20°
x = 35°
Explanation:
Equation's:
1) 2y + x + 105° = 180°
2) 3x + x + 2y = 180°
Make y subject in equation 2
3x + x + 2y = 180
4x + 2y = 180
2y = 180 - 4x
y = 90 - 2x
Insert this into equation 1
2(90 - 2x) + x + 105° = 180°
180 - 4x + x + 105 = 180
-3x = -105
x = 35°
Find value of y
y = 90 - 2x
y = 90 - 2(35)
y = 20°
Answer:
<h2>Required set will be :-</h2><h3>(9,15)</h3>
Answer:
The 13th term is 81<em>x</em> + 59.
Step-by-step explanation:
We are given the arithmetic sequence:
And we want to find the 13th term.
Recall that for an arithmetic sequence, each subsequent term only differ by a common difference <em>d</em>. In other words:
Find the common difference by subtracting the first term from the second:
Distribute:
Combine like terms. Hence:
The common difference is (7<em>x</em> + 5).
To find the 13th term, we can write a direct formula. The direct formula for an arithmetic sequence has the form:
Where <em>a</em> is the initial term and <em>d</em> is the common difference.
The initial term is (-3<em>x</em> - 1) and the common difference is (7<em>x</em> + 5). Hence:
To find the 13th term, let <em>n</em> = 13. Hence:
Simplify:
The 13th term is 81<em>x</em> + 59.
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 25235
For the alternative hypothesis,
µ > 25235
This is a right tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 100,
Degrees of freedom, df = n - 1 = 100 - 1 = 99
t = (x - µ)/(s/√n)
Where
x = sample mean = 27524
µ = population mean = 25235
s = samples standard deviation = 6000
t = (27524 - 25235)/(6000/√100) = 3.815
We would determine the p value using the t test calculator. It becomes
p = 0.000119
Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.
