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2 years ago

11

Round to the nearest hundredth (Geometric Series problem)

1 answer:

WITCHER [35]2 years ago

4 0

Answer:

15.98

Step-by-step explanation:

∑₁¹⁰ $8(\frac{1}{4} )^{n-1}$

$=8[1+\frac{1}{4} +\frac{1}{16} + ........ +t_{10}]$

We have to use the formula of sum of a G.P. series with common ratio r < 1.

If the first term is a, the common ratio is r and the number of terms is n then the sum [a + ar + ar² + ar³ + ......... up to n terms] is given by

$[a \times\frac{1-r^{n} }{1-r}]$

$= 8[1 \times \frac{1-\frac{1}{2^{10} } }{1-\frac{1}{2} } ]$

$= 15.98$ (Answer)

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Find the sum of the first<br>20 odd numbers.

konstantin123 [22]

Answer:

Hi there.....

Step-by-step explanation:

This is ur answer......

1, 3, 5, 7, 9, 11, 13, 15.................39

Therefore the total sum = 400

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3 years ago

Read 2 more answers

What is the least common multiple of 8 and 10

Lynna [10]

40 is your answer. Because, you need the lowest multiple so we have to find it by making out a chart of their multiples

8, 16,24,32,40

10,20,30,40

The first same multiple they make is 40

3 0

3 years ago

Read 2 more answers

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago

On the planet Tott, there are 7 rews to every 6 hopts. If farmer Boondin has 40 hopts on her frenk farm, how many rews are on th

Anestetic [448]

Answer:

uwuu

Step-by-step explanation:

ok the reason y iput uwu because uwu

4 0

2 years ago

Read 2 more answers

83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

$\chi^2_{\alpha/2}=104.139$

$\chi^2_{1- \alpha/2}=62.132$

The confidence interval is given by:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

5 0

2 years ago