11+2 to the 3rd power ×5 using PEMDAS?

2 answers:

8 0

11+2^3x5

You always start with parenthesis but there aren't any so you go to exponents. 2 to the third power is 2 x 2 x 2 which is 8. so now you will have 11+8x5. Next is either multiply or divide left to right. There's only multiplication so you have to multiple 8 x 5 which is 40. now all you're left with is 11 + 40. You have to add or subtract from left to right. Since, there's only adding, 11 + 40 = 51

allochka39001 [22]3 years ago

6 0

58 - its simple- Hope I helped :)

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g (15 points) Suppose 42 out of 600 rats exposed to a potential carcinogen develop tumors. A control group of 350 rats not expos

vazorg [7]

Answer:

a) The relative risk is 1.8 $\overline{846153}$

b) The attributable risk is $0.69 \overline{047619}$

c) There is a relationship between exposure and tumor risk

Step-by-step explanation:

The number of exposed rats that develop tumors, a = 42

The number of rats exposed to the carcinogen = 600 rats

The number of exposed rats that did not develop tumors, b = 600 - 42 = 558

The number of not exposed rats in the control group = 350 rats

The number of rats that develop tumors in the control group, c = 13 tumors

The number of not exposed rats that did not develop tumors, d = 350 - 13 = 337 rats

a) The relative risk, RR = a/(a + b)/(c/(c + d))

∴ RR = (42/(42 + 558))/(13/(13 + 337)) = 49/26 = 1.8 $\overline{846153}$

b) The attributable risk = (a - c)/a

∴ The attributable risk = (42 - 13)/42 = $0.69 \overline{047619}$

c) The odds ratio = (a·b)/(c·d)

∴ The odds ratio = 42 × 558/(13 × 337) = 23439/4381 ≈ 5.35

Given that the result of attributable risk is positive, there is an indication that there is a higher probability to develop tumor when exposed to the potential carcinogen, therefore, there is a relationship between exposure and tumor risk

7 0

3 years ago

Given that triangle ABC is a right triangle, select a set of possible side length measures.

ad-work [718]

You didn't supply any rules or constraints so.... 1 in., 1 in., and 1 in.

You can use the Pythagorean Theorem to check if the side lengths are appropriate.

$a^2 + b^2 = c^2$