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ArbitrLikvidat [17]
3 years ago
5

¿Cuánto es 15% de 3 000?​

Mathematics
2 answers:
ra1l [238]3 years ago
6 0
La repuesta es 450!
kati45 [8]3 years ago
4 0
la Repuesta es 450!
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Which expression is equivalent to 7 (x + 4)? 28 x 7 (x) + 7 (4) 7 (x) + 4 11 x
Rama09 [41]

Answer:

i'm guessing you're asking which expression is the same as 7(x+4)

i don't know what the other numbers listed are suppose to mean, but when expanded, 7(x+4)=7x+28

anything added together, multiplied together, etc, that equals 7x+28 is your answer

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3 years ago
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A plus 5 equals negative 5a plus 5
jenyasd209 [6]
A+5=5a+5   Subtract a from each side
5=5a+5       Subtract 5 from each side
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--- ---
5   5
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A jar of peanut butter and a jar of jam cost $10.20 in total. the jar of peanut butter costs $10.00 more than the jar of jam. ho
Olegator [25]
Well this is pretty simple. So the first thought is that the peanut butter would be 10$ and the jam would be 0.20$, however, the peanut butter would not be 10$ more. Instead, subtract the 10$ from the total, which gives you 0.20$, and then divide that by two. Now you have 0.10$ for each, along with another 10$ for the peanut butter. The peanut butter would be $10.10, and the jam would be 0.10$ (that's pretty cheap!).
4 0
3 years ago
... What is P(A/B)?
Orlov [11]
Uuuuhhhhhh I am sooooo confused as I haven’t even learned this concept yet
3 0
3 years ago
An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30, de
Lelechka [254]

Answer:

(a) P(X=10) = 0.2070

(b) P(X\geq 10) = 0.3798

Step-by-step explanation:

Note that in this problem we have an initial population N = 50, of which 30 fulfill a certain characteristic "m" (belong to the second section). Then, from the population N, a sample of size n = 15 is selected and it is desired to know how many comply with the desired characteristic (second section).

So

Let X be the number of projects in the second section, then X is a discrete random variable that can be modeled by a hypergeometric distribution.

(a)

Therefore, to answer question (a) we use the following equation presented in the attached image:

Where:

N = 50\\m = 30\\n = 15\\X = 10

Then:

P (X = 10) = \frac{\frac{30!}{10!(30-10)!}\frac{20!}{5!(20-5)!}}{\frac{50!}{15!(50-15)!}}\\\\\\P(X=10) = 0.2070

(b)

For part (b) we have:

P(X\geq10) = 1-P(X

P(X\geq 10) = 1-0.6202 \\\\P(X\geq 10) = 0.3798

4 0
3 years ago
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