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3 years ago

Why should you look out for a pig that knows karate?

Mathematics

2 answers:

prohojiy [21]3 years ago

8 0

We should look out for a pig that knows karate because it might give up a pork chop

<h3>Further explanation </h3>

According to the dictionary, pig is an omnivorous domesticated hoofed mammal with sparse bristly hair and a flat snout for rooting in the soil, kept for its meat. Whereas the karate pig is like fictional character of pig, where the pig is able to do karate. Karate is a martial art developed in the Ryukyu Kingdom. Karate means "empty hand."

Karate pig can do the pork chop, so we call a pig that does karate as the pork chop. Pork chop is the meat chops made from pork. It is a loin cut taken perpendicularly to the pig spine and it usually contain a rib or part of a vertebra. It are not processed than other cuts of pork.

This is an old joke :P

There are recipe to make pork chop. First preheat oven to 350 degrees F. In a small bowl, blend soy sauce, vegetable oil, sauce, lemon juice, brown sugar, and ketchup. Then lace pork chops in a medium baking dish, and spread with 1/2 the sauce. At the end, bake pork chops 30 minutes in the preheated oven.

<h3>Learn more</h3>

Learn more about pig brainly.com/question/9875797
Learn more about karate brainly.com/question/10809353
Learn more about chop brainly.com/question/803560

<h3>Answer details</h3>

Grade: 5

Subject: Math

Chapter: animals

Keywords: pig, animals, karate, look out, chop

kipiarov [429]3 years ago

5 0

It might give you a pork chop....
Math teachers usually give this riddle for you tio make a sentence from correct letters....
Hope I could help :)<span />

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Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

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A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

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 weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago