I would like some assistance pls

Find the 7th term of the geometric sequence whose common ratio is 1/2 and whose first term is 5

Alchen [17]

Answer:

Step-by-step explanation:

We are given that:

First term (a) = 5

common ratio (r) =1/3

So, the t of 7 term is 5*1/729=5/729

8 0

3 years ago

Q4.

goblinko [34]

The coding of the statistic is used to make it easier to work with the large sunshine data set

The mean of the sunshine is 3.05 $\overline 6$

The standard deviation is approximately 18.184

Reason:

The given parameters are;

The sample size, n = 3.

∑x = 947

Sample corrected sum of squares, Sₓₓ = 33,065.37

The mean and standard deviation = Required

Solution:

$Mean, \ \overline x = \dfrac{\sum x_i}{n}$

The mean of the daily total sunshine is therefore;

$Mean, \ \overline x = \dfrac{947}{30} \approx 31.5 \overline 6$

$s = \dfrac{x}{10 } - \dfrac{1}{10}$

$E(s) = \dfrac{Ex}{10 } - \dfrac{1}{10}$

$E(s) = \dfrac{31.5 \overline 6}{10 } - \dfrac{1}{10} = 3.05 \overline 6$

The mean ≈ 3.05 $\overline 6$

Alternatively

, $The \ mean \ of \ the \ daily \ total \ sunshine, \, s = \dfrac{31.5 \overline 6 - 1}{10 } = 3.05\overline 6$

The mean of the daily total sunshine, $\overline s$ ≈3.05 $\overline 6$

$Var(s) = Var \left(\dfrac{x}{10 } - \dfrac{1}{10} \right)$

$Var(s) = \left(\dfrac{1}{10}\right)^2 \times Var \left(x \right)$

Therefore;

$Var(s) = \left(\dfrac{1}{10}\right)^2 \times 33,065.37 = 330,6537$

Therefore;

$s = \sqrt{330.6537} \approx 18.184$

The standard deviation, $s_s$ ≈ 18.184

Learn more about coding of statistic data here:

brainly.com/question/14837870

3 0

2 years ago

1

NikAS [45]

9514 1404 393

Answer:

1.0 N

Step-by-step explanation:

When the length increases by the factor 120/80 = 3/2, the force required is multiplied by the inverse of this: 2/3.

The required force is ...

(2/3)(1.5 N) = 1.0 N

5 0

3 years ago

Find the bank discount and proceeds using ordinary interest for a loan to michelle anders for 7,200 at 8.25% annual simple inter

FromTheMoon [43]

Answer:

$145.5

Step-by-step explanation:

Period, t = August 8.to November 8 = 3 months

Simple interest formula :

A = P(1 + rt)

A = final amount =. 7200

7200 = P(1 + (0.0825 * 3/12))

7200 = P(1 + 0.020625)

7200 = P(1.020625)

7200 = 1.020625P

P = 7200 / 1.020625

P = 7054.5009

Bank discount :

A - P

7200 - 7054.5009

= $145.4991

= $145.5

6 0

3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How

DedPeter [7]

Answer:

a) $Q(t) = 180e^{-0.023t}$

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-30r}$

$e^{-30r} = 0.5$

Applying ln to both sides of the equality.

$\ln{e^{-30r}} = \ln{0.5}$

$-30r = \ln{0.5}$

$r = \frac{\ln{0.5}}{-30}$

$r = 0.023$

So

$Q(t) = Q(0)e^{-0.023t}$

180-mg sample, so Q(0) = 180

$Q(t) = 180e^{-0.023t}$

(b) How much of the sample remains after 120 years?

This is Q(120).

$Q(t) = 180e^{-0.023t}$

$Q(120) = 180e^{-0.023*120}$

$Q(120) = 11.4$

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

$Q(t) = 180e^{-0.023t}$

$1 = 180e^{-0.023t}$

$e^{-0.023t} = \frac{1}{180}$

$e^{-0.023t} = 0.00556$

Applying ln to both sides

$\ln{e^{-0.023t}} = \ln{0.00556}$

$-0.023t = \ln{0.00556}$

$t = \frac{\ln{0.00556}}{-0.023}$

$t = 225.8$

225.8 years.

8 0

3 years ago

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I would like some assistance pls​

I would like some assistance pls