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3 years ago

Can you help me calculate the angles from this kite in the attachment?AY =, XZ =,

Mathematics

1 answer:

Setler [38]3 years ago

8 0

Answer:

AY=8, XZ=20

Step-by-step explanation:

WY=WA+AY, plug in values: 14=6+AY, subtract 6 from both sides: 8=AY, AY=8cm

ZA=10cm and WY bisects XZ at A, so XA=ZA and XZ = XA+ZA = 2ZA =2*10 = 20 cm

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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