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dmitriy555 [2]
3 years ago
14

Pls help me I have other homework to do. Also it detects if it’s right or wrong:(

Mathematics
1 answer:
Lostsunrise [7]3 years ago
4 0

Answer:

480-1-x=330 i think

Step-by-step explanation:

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8 2/1

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3 years ago
Find two factors of the first number such that their product is the first and the sum is the second for 51,20
anygoal [31]

The factors are 17 and 3.

The given conditions are two factors of the first number, 51 such that their product is 51 and the sum is the second number 20.
let say, the factors are x and y.

therefore, we can write, x.y = 51 ... (1)  and x + y = 20 .... (2)

Putting  the value of x from equation (2) in  equation (1) we get,

(20 - y).y = 51

or, 20y - y² = 51

or, y² - 20y + 51 = 0

or, y = 17, 3

then x = 3,17

Therefore, The factors are 17 and 3.

Learn more about factors here:

brainly.com/question/24182713

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6 0
2 years ago
What is the atomic radius of a BCC crystal structure?​
Semenov [28]

Answer:

Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol.

8 0
3 years ago
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Scores on a statewide standardized test are normally distributed with a mean of 12.89 and a standard deviation of 1.95. Certific
Ulleksa [173]

Answer:

Marcus score is 13.7, the mean is 12.89, and the standard deviation is 1.95, so I’ll do ( 13.7 – 12.89)/ 1.95, Since Marcus' z-score is 0.415. The area to the left of the z-score is 65.9% so therefore Marcus did scored better than 65.9% of the students. So Marcus needs to score 98% better of all the other students in order for him to receive an actual certificate.

Step-by-step explanation:

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3 years ago
A scientist inoculates mice, one at a time, with a disease germ until he finds 2 that have contracted the disease. if the probab
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<span>There are several possible events that lead to the eighth mouse tested being the second mouse poisoned. There must be only a single mouse poisoned before the eighth is tested, but this first poisoning could occur with the first, second, third, fourth, fifth, sixth, or seventh mouse. Thus there are seven events that describe the scenario we are concerned with. With each event, we want two particular mice to become diseased (1/6 chance) and the remaining six mice to remain undiseased (5/6 chance). Thus, for each of the seven events, the probability of this event occurring among all events is (1/6)^2(5/6)^6. Since there are seven of these events which are mutually exclusive, we sum the probabilities: our desired probability is 7(1/6)^2(5/6)^6 = (7*5^6)/(6^8).</span>
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