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3 years ago

What is the 10th term of the geometric sequence 400, 200, 100...?

Mathematics

1 answer:

Vsevolod [243]3 years ago

4 0

ANSWER

$a_ {10} = \frac{25}{32}$

EXPLANATION

The given geometric sequence is

400, 200, 100...

The first term is

$a_1=400$

The common ratio is

$r = \frac{200}{400} = \frac{1}{2}$

The nth term is

$a_n=a_1( {r}^{n - 1} )$

We substitute the known values to get;

$a_n=400( \frac{1}{2} )^{n - 1}$

$a_ {10} =400( \frac{1}{2} )^{10 - 1}$

$a_ {10} =400( \frac{1}{2} )^{9}$

$a_ {10} = \frac{25}{32}$

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Solve the expression below, when x = 8 (x - 4) + 12

LenaWriter [7]

Answer:

Step-by-step explanation:

Lets solve this step by step,

First we insert 8 in x's place:

(8 - 4) + 12

Now we solve:

8 - 4 = 4

(4) + 12

4 + 12 = 16

Answer: 16

Hope this Helps

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3 years ago

Lan Ventor designs a new board game and offers the design to Paddington Games p.l.c.

Andrej [43]

In order to make the offer attractive such that it would earn £25,000 for Ian Vector, Paddington Games would have to sell 460 games.

The game can be sold for £25,000 or for £2,000 and then a fee of £50 for every game sold.

In order for the amount to be the same, the amount from games sold will have to equal the difference between the £25,000 and the £2,000.

Difference is:

= 25,000 - 2,000

= £23,000

The number of games to be sold is:

= Difference / Amount per game

= 23,000 / 50

= 460 games

In conclusion, 460 games need to be sold to make the offer attractive.

Find out more at brainly.com/question/2865277.

5 0

2 years ago

A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two

Vikentia [17]

Answer:

the probability of no defects in 10 feet of steel = 0.1353

Step-by-step explanation:

GIven that:

A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two.

Let consider β to be the average value for defecting

So;

β = 2

Assuming Y to be the random variable which signifies the anticipated number of defects in a 10-foot segment of this roll.

Thus, y follows a poisson distribution as number of defect is infinite with the average value of β = 2

i.e

$Y \sim P( \beta = 2)$

the probability mass function can be represented as follows:

$\mathtt{P(y) = \dfrac{e^{- \beta} \ \beta^ \ y}{y!}}$

where;

y = 0,1,2,3 ...

Hence, the probability of no defects in 10 feet of steel

y = 0

$\mathtt{P(y =0) = \dfrac{e^{- 2} \ 2^ \ 0}{0!}}$

$\mathtt{P(y =0) = \dfrac{0.1353 \times 1}{1}}$

P(y =0) = 0.1353

4 0

3 years ago

Ava's birthday cake is about 11 7/10 cm long. her candles stick up from the cake an additional 5 4/10 cm. what was the total hig

Virty [35]

Total height of the cake and the candles is 17.1 cm

Step-by-step explanation:

Step 1: Convert the mixed fractions into decimal expressions.

11 7/10 cm = 117/10 = 11.7 cm

5 4/10 cm = 54/10 = 5.4 cm

Step 2: Find total height of the cake and the candles

Total height = 11.7 + 5.4 = 17.1 cm

3 0

3 years ago

Read 2 more answers

Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Batt

-Dominant- [34]

Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is tested if it does not outlast by more than 2 hours, that is, the subtraction is not more than 2:

$H_0: \mu_B - mu_A \leq 2$

At the alternative hypothesis, it is tested if it outlasts by more than 2 hours, that is:

$H_1: \mu_B - \mu_A > 2$

The sample means are: $\mu_A = 8.65, \mu_B = 11.23$

The standard deviations for the samples are $s_A = s_B = 0.67$

Hence, the standard errors are:

$s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934$

The distribution of the difference has mean and standard deviation given by:

$\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58$

$s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 2$ is the value tested at the hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{2.58 - 2}{0.2735}$

$t = 2.12$

The critical value for a right-tailed test, as we are testing if the subtraction is greater than a value, with a 0.05 significance level and 12 + 12 - 2 = 22 df is given by $t^{\ast} = 1.71$

Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630

7 0

3 years ago