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Helga [31]
1 year ago
7

PLEASE HELP ASAP Eva is at a sushi restaurant. She ordered 2 pieces of squid for a total of $3.50, 1 piece of eel for $3.25, 3 p

ieces of tuna for a total of $6.75, and 4 pieces of crab for $8.00. Which list shows the unit cost of each pieces of sushi from least to greatest? A. squid, crab, tuna, eel B. eel, squid, tuna, crab C. squid, eel, crab, tuna D. eel, tuna, crab, squid
Mathematics
1 answer:
mrs_skeptik [129]1 year ago
4 0

Answer:

I do not have enough time to put them in order if you don't mind but i can list the prices.

Step-by-step explanation:

each piece of squid is 1.25

each piece of tuna is 3.25

each piece of crab is 2.00

hope this helps <3

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1) We have that there are (52,3) ways to pick 3 cards out of 52 cards. Also, there are (4,3) ways to pick 3 kings out of the 4 total available kings in the deck.  In essence, we need to have one of those ways to be the selected 3 cards. Hence, the probability is the ration (4,3)/(52,3). Computing this:
P=\frac{1*2*3*4}{(1*2*3)*1} /  \frac{52*51*50}{1*2*3} = \\  \frac{4*1*2*3}{52*51*50}=0.0181%
The probability is very low but not negligible.

2) The Pascal triangle defines a recursive relationship. Hence, we would need to calculate all the binomial coefficients up to 51. Thus, it is not at all practical to use the Pascal Triangle to calculate the ways. It is easier to do the direct computation.
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The total cost of a jacket and a pair of shoes was $99.15. if the price of the jacket was $5.81 less than the pair of shoes, wha
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An Article in the Journal of Sports Science (1987, Vol. 5, pp. 261-271) presents the results of an investigation of the hemoglob
ValentinkaMS [17]

Answer:

The 95% confidence interval for the population variance is \left[0.219, \hspace{0.1cm} 0.807\right]\\\\

The 95% confidence interval for the population mean is \left [15.112, \hspace{0.3cm}15.688\right]

Step-by-step explanation:

To solve this problem, a confidence interval of (1-\alpha) \times 100% for the population variance will be calculated.

$$Sample variance: $S^2=(0.6152)^2$\\Sample size $n=20$\\Confidence level $(1-\alpha)\times100\%=95\%$\\$\alpha: \alpha=0.05$\\$\chi^2$ values (for a 95\% confidence and n-1 degree of freedom)\\$\chi^2_{\left (1-\frac{\alpha}{2};n-1\right )}=\chi^2_{(0.975;19)}=8.907\\$\chi^2_{\left (\frac{\alpha}{2};n-1\right )}=\chi^2_{(0.025;19)}=32.852\\\\

Then, the (1-\alpha) \times 100% confidence interval for the population variance is given by:

\left [\frac{(n-1)S^2}{\chi^2_{\left (\frac{\alpha}{2};n-1\right )}}, \hspace{0.3cm}\frac{(n-1)S^2}{\chi^2_{\left (1-\frac{\alpha}{2};n-1\right )}} \right ]\\\\Thus, the 95% confidence interval for the population variance is:\\\\\left [\frac{(19-1)(0.6152)^2}{32.852}, \hspace{0.1cm}\frac{(19-1)(0.6152)^2}{8.907} \right ]=\left[0.219, \hspace{0.1cm} 0.807\right]\\\\

On other hand,

A confidence interval of (1-\alpha) \times 100% for the population mean will be calculated

$$Sample mean: $\bar X=15.40$\\Sample variance: $S^2=(0.6152)^2$\\Sample size $n=20$\\Confidence level $(1-\alpha)\times100\%=95\%$\\$\alpha: \alpha=0.05$\\T values (for a 95\% confidence and n-1 degree of freedom) T_{(\alpha/2;n-1)}=T_{(0.025;19)}=2.093\\\\$Then, the (1-\alpha) \times 100$\% confidence interval for the population mean is given by:\\\\

\\left[ \bar X - T_{(\alpha/2;n-1}\sqrt{\frac{\S^2}{n}}, \hspace{0.3cm}\bar X + T_{(\alpha/2;n-1}\sqrt{\frac{\S^2}{n}} \right ]\\\\Thus, the 95\% confidence interval for the population mean is:\\\\\left [15.40 - 2.093\sqrt{\frac{(0.6152)^2}{19}}, \hspace{0.3cm}15.40 + 2.093\sqrt{\frac{(0.6152)^2}{19}} \right ]=\left [15.112, \hspace{0.3cm}15.688\right] \\\\

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2 years ago
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