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3 years ago

Find the 39th term of the arithmetic sequence

Mathematics

1 answer:

myrzilka [38]3 years ago

8 0

Answer:

-142

Step-by-step explanation:

You seem to have an arithmetic sequence with first term -66 and common difference (-68-(-66)) = -2. The general term of such a sequence is given by ...

an = a1 +d(n -1) . . . . . for first term a1 and common difference d

We want to evaluate this equation for a1 = -66, d = -2, n = 39:

a39 = -66 -2(39 -1) = -66 -76 = -142

The 39th term is -142.

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Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

$\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}$

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

$x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13$

$x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88$

$x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63$

$x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38$

$x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13$

So, the table becomes:

$\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}$

Next, calculate the mean

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}$

$\bar x = \frac{1845.73}{146}$

$\bar x = 12.64$

Next, the sample variance is:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}$

So, we have:

$\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}$

$\sigma^2 = \frac{2393.6875}{145}$

$\sigma^2 = 16.51$

The sample standard deviation is:

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{16.51}$

$\sigma = 4.06$

6 0

2 years ago

Solve for x. Round to the nearesr tenth of a degree if necessary.

Ipatiy [6.2K]

Answer:

im not sure but i got 30.1

Step-by-step explanation:

$\cos(x) = \frac{45}{52} \\ x = 30.1$

7 0

3 years ago

Help me w this problem please !!!

jeka94

Answer:

it depends in u

and I will be there at noon and I will get back to Live in the morning and I Challenge u Day my

3 0

2 years ago

Can someone please help with 15?

natima [27]

The answer would be c

8 0

3 years ago

The sum of three numbers in <br> g.p. is 21 and the sum of their squares is 189. find the numbers.

sashaice [31]

Let the three gp be a, ar and ar^2
a + ar + ar^2 = 21 => a(1 + r + r^2) = 21 . . . (1)
a^2 + a^2r^2 + a^2r^4 = 189 => a^2(1 + r^2 + r^4) = 189 . . . (2)
squaring (1) gives
a^2(1 + r + r^2)^2 = 441 . . . (3)
(3) ÷ (2) => (1 + r + r^2)^2 / (1 + r^2 + r^4) = 441/189 = 7/3
3(1 + r + r^2)^2 = 7(1 + r^2 + r^4)
3(r^4 + 2r^3 + 3r^2 + 2r + 1) = 7(1 + r^2 + r^4)
3r^4 + 6r^3 + 9r^2 + 6r + 3 = 7 + 7r^2 + 7r^4
4r^4 - 6r^3 - 2r^2 - 6r + 4 = 0
r = 1/2 or r = 2
From (1), a = 21/(1 + r + r^2)
When r = 2:
a = 21/(1 + 2 + 4) = 21/7 = 3
Therefore, the numbers are 3, 6 and 12.

3 0

3 years ago