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3 years ago

X + y = 0

Mathematics

1 answer:

IrinaVladis [17]3 years ago

7 0

Answer:

(-3,6) (1,8) (2,10) and so on.

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9/20

NISA [10]

For this case we must find the solution of the following quadratic equation:

$x ^ 2 + 5x + 7 = 0$

Where:

$a = 1\\b = 5\\c = 7$

Then, the solution is given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Substituting the values:

$x = \frac {-5 \pm \sqrt {5 ^ 2-4 (1) (7)}} {2 (1)}\\x = \frac {-5 \pm \sqrt {25-28}} {2}\\x = \frac {-5 \pm \sqrt {-3}} {2}$

By definition we have: $i ^ 2 = -1$

$x = \frac {-5 \pm \sqrt {3i ^ 2}} {2}\\x = \frac {-5 \pm i \sqrt {3}} {2}$

Thus, we have two complex roots.

Answer:

The equation has no real roots.

4 0

4 years ago

Convert the following degree measure to radian measure. 90 degrees

Ray Of Light [21]

90 degrees converted to radians is pi/2 or π/2 radians.

4 0

3 years ago

A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted is incorrectly received wit

m_a_m_a [10]

Answer:

The probability that the message will be wrong when decoded is 0.05792

Step-by-step explanation:

Consider the provided information.

To reduce the chance or error, we transmit 00000 instead of 0 and 11111 instead of 1.

We have 5 bits, message will be corrupt if at least 3 bits are incorrect for the same block.

The digit transmitted is incorrectly received with probability p = 0.2

The probability of receiving a digit correctly is q = 1 - 0.2 = 0.8

We want the probability that the message will be wrong when decoded.

This can be written as:

$P(X\geq3) =P(X=3)+P(X=4)+P(X=5)\\P(X\geq3) =\frac{5!}{3!2!}(0.2)^3(0.8)^{2}+\frac{5!}{4!1!}(0.2)^4(0.8)^{1}+\frac{5!}{5!}(0.2)^5(0.8)^0\\P(X\geq3) =0.05792$

Hence, the probability that the message will be wrong when decoded is 0.05792

4 0

3 years ago

A Food Marketing Institute found that 39% of households spend more than $125 a week on groceries. Assume the population proporti

Anuta_ua [19.1K]

Answer:

0.6210

Step-by-step explanation:

Given that a Food Marketing Institute found that 39% of households spend more than $125 a week on groceries

Sample size n =87

Sample proportion will follow a normal distribution with p =0.39

and standard error = $\sqrt{\frac{0.39(1-0.39)}{87} } \\=0.0523$

the probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41

= $P(0.29$

There is 0.6210 probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41

7 0

4 years ago

HELP ME ASAP PLS.......

Sergio [31]

Answer:

Step-by-step explanation: 32

3 0

3 years ago