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Sav [38]
3 years ago
14

Find the center of the ellipse. x2 + 4y2 – 10x – 40y + 121 = 0​

Mathematics
2 answers:
EleoNora [17]3 years ago
5 0

Answer:

i dont what an ellipse is but here's the answer:

8x + 32y = 121

xz_007 [3.2K]3 years ago
4 0

Answer:

123‐10×40y=0

10×+40y=123

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Answer:

$0.75

Step-by-step explanation:

28.50-24=$4.50

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6 0
3 years ago
3=c/5+2... how do I solve for c??
expeople1 [14]
Primeiro Você IRA Dividir o c (APENAS Colocar 5 + 2 embaixo do c). Depois faça a soma 5 + 2 = 7. Então faça o mmc (mínimo múltiplo comum) entre 7 e 1(o 1 é invisível, mas continua estando embaixo do 3). Deu 7 o mmc. Transforme seus números em frações, com o denominador 7. Transforme os números em frações o 3 virá 21, pois você divide em baixo e multiplica em baixo, e o c continua normal, pois já estava em baixo de 7. Como é para descobrir uma incógnita você tira os denominadores e ficará 21 = c.

3 = c/5 + 2
3 = c/7
21/7 = c/7
21 = c
4 0
3 years ago
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Help i gotta get it turned in soon​
Katen [24]

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Step-by-step explanation:

4 0
3 years ago
A teacher teaches two classes with 8 students each. Each student has a 95% chance of passing their class independent of the othe
dsp73

Answer:

0.4466 = 44.66% probability that, in exactly one of the two classes, all 8 students pass.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they pass, or they do not. The probability of an student passing is independent of other students, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability that all students pass in a class:

Class of 8 students, which means that n = 8

Each student has a 95% chance of passing their class independent of the other students, which means that p = 0.95

This probability is P(X = 8). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{8,8}.(0.95)^{8}.(0.05)^{0} = 0.6634

Find the probability that, in exactly one of the two classes, all 8 students pass.

Two classes means that n = 2

0.6634 probability all students pass in a class, which means that p = 0.6634.

This probability is P(X = 1). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{2,1}.(0.6634)^{1}.(0.3366)^{1} = 0.4466

0.4466 = 44.66% probability that, in exactly one of the two classes, all 8 students pass.

3 0
3 years ago
Which side does the right angle always point to?<br><br>​
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Answer:

c the hypotnuse of the triangle

plz give brainliest

Step-by-step explanation:

6 0
3 years ago
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