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3 years ago

Help me solve: 3.14•8^2•23 And round to nearest tenth if necessary

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

7 0

Answer:

4622.1

Step-by-step explanation:

We can use PEMDAS for this

in PEMDAS exponents comes 2nd and in the expression we have exponents so lets solve that first

8^2=64

next is multiplication

64*3.14=200.96

200.96*23=4622.08

to round to the nearest tenth we need to look right of the tenths place

since there is an "8" we can round up to "1"

4622.1

Hope this helps

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98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

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