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3 years ago

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At a local high school, the student population is growing at 12% a year. If the original population was 242 students, how long w

ill it take the population to reach 300 students? Round to the nearest tenth of a year.

1 answer:

uranmaximum [27]3 years ago

8 0

Answer: 2 years

Step-by-step explanation:

The exponential growth function is given by :-

$y=A(1+r)^x$ (i)

, where A = initial value , r = rate of growth and x= time period.

As per given ,

A= 242

r= 12% = 0.12

To find : t when y= 300.

Put all the values in (i)

$300=242(1+0.12)^x\\\\\Rightarrow\ \dfrac{300}{242}=(1.12)^x\\\\\Rightarrow\ 1.23967=(1.12)^x$

Taking log on both sides , we get

$\log (1.2396) = t \log (1.12)\\\\\Rightarrow\ 0.09328=t(0.049218)\\\\\Rightarrow t=\dfrac{0.09328}{0.049218}=\approx2$

hence, it will take 2 years.

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Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15. N an 1 4 2 −12 3 36 t

Rzqust [24]

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is $\sum_{n=4}^{15} 5(-2)^{n-1}$$$

<h3>What is sequence ?</h3>

Sequence is collection of numbers with some pattern .

Given sequence

$a_{1}=5\\\\a_{2}=-10\\\\\\a_{3}=20$

We can see that

$\frac{a_1}{a_2}=\frac{-10}{5}=-2\\$

and

$\frac{a_2}{a_3}=\frac{20}{-10}=-2\\$

Hence we can say that given sequence is Geometric progression whose first term is 5 and common ratio is -2

Now $n^{th}$ term of this Geometric progression can be written as

$T_{n}= 5\times(-2)^{n-1}$

So summation of 15 terms can be written as

$\sum_{n=4}^{15} T_{n}\\\\$\\$\sum_{n=4}^{15} 5(-2)^{n-1}$$$

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is $\sum_{n=4}^{15} 5(-2)^{n-1}$$$

To learn more about Geometric progression visit : brainly.com/question/14320920

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3 years ago

One True Love? A survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each pe

sertanlavr [38]

Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of people who disagree with the statement = $\frac{1812}{2625}$ = 0.69

n = sample of respondents = 2625

p = population proportion of people who disagree with statement

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

<u>So, 99% confidence interval for the population proportion, p is ;</u>

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<u>99% confidence interval for p</u> = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ , $0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ ]

= [0.667 , 0.713]

Therefore, 99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

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