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3 years ago

place an x in the table to show the percent decrease from the first number to the sound number. GIVING BRAINLIEST!!!

Mathematics

2 answers:

Mamont248 [21]3 years ago

8 0

Answer:

Step-by-step explanation:

1. 150 to 132 is a decrease of 12%

2. 275 to 242 is a decrease of 12%

3. 300 to 255 is a decrease of 15%

Thanks For the Brainliest!

Viefleur [7K]3 years ago

8 0

Answer:

1. 150 - 12% = 132

2. 275 - 12% = 242

3. 300 - 15% = 255

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An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago

Jojo bear bought 2 juices for 2.95 per juice and three bags of chips for 1.20 each. How much did he spend in total. Show work

stealth61 [152]

He spent $9.50 in total. To find this answer, you write the equation 2(9.95) + 3(1.20). You would distribute getting 5.90 + 3.60= $9.50.

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3 years ago

Roberto debe leer un libro que contiene 340 páginas en cuatro días para lo cual planea la siguiente distribución de lectura: el

Murrr4er [49]

Answer: le quedan leer 295 páginas

Step-by-step explanation:

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3 years ago

Is it greater than, Less than, or equal to?

frosja888 [35]

The answer is greater than

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3 years ago

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In a group of 120 recruits at the police academy 80 are college graduates and 25 are army veterans. If 15 of the recruits are bo

Gre4nikov [31]

Answer:

the number of recruits neither be college graduate and nor army veterans is 30

Step-by-step explanation:

The computation of the number of recruits neither be college graduate and nor army veterans is as follows;

= Total recruits - (college students + army veterans - both college graduates & army veterans)

= 120 recruits - (80 + 25 - 15)

= 120 recruits - 90

= 30

hence, the number of recruits neither be college graduate and nor army veterans is 30

4 0

3 years ago