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2 years ago

Amzol thinks that (x+5)^2=x^2+25 for all values of x. Is Amzol right?

Mathematics

1 answer:

Viefleur [7K]2 years ago

5 0

Answer:

The answer is x=0 so probably he is

Step-by-step explanation:

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Assume that different groups of couples use the XSORT method of gender selection and each couple gives birth to one baby. The XS

olganol [36]

Answer:

Mean and Standard deviation for the numbers of girls in groups of 36 births are 18 and 3 respectively.

Step-by-step explanation:

We are given that he X SORT method is designed to increase the likelihood that a baby will be a girl, but assume that the method has no effect, so the probability of a girl is 0.5.

Now consider a group consisting of 36 couples.

The above situation can be represented through binomial distribution;

$P(X=r)=\binom{n}{r} \times p^{r} \times (1-p)^{n-r} ;x=0,1,2,3,.....$

where, n = number of trials (samples) taken = 36 couples

r = number of success

p = probability of success which in our question is probability

of a girl, i.e.; p = 0.5

Let X = Numbers of girls in groups of 36 births

So, X ~ Binom(n = 36, p = 0.5)

Now, mean for the numbers of girls in groups of 36 births is given by;

Mean, E(X) = $n \times p$ = $36 \times 0.5$ = 18

Also, standard deviation for the numbers of girls in groups of 36 births is given by;

Standard deviation, S.D.(X) = $\sqrt{n\times p\times (1-p)}$

= $\sqrt{36\times 0.5\times (1-0.5)}$

= 3

3 0

3 years ago

write a polynomial function of least degree with integral coefficients that has the given zeros. -(1/3), -i

inessss [21]

Answer:

$f(x)=3x^3+x^2+3x+1$

Step-by-step explanation:

If a real number $-\frac{1}{3}$ is a zero of polynomial function, then

$x-\left(-\dfrac{1}{3}\right)=x+\dfrac{1}{3}$

is the factor of this function.

If a complex number $-i$ is a xero of the polynomial function, then the complex number $i$ is also a zero of this function and

$x-(-i)=x+i\ \text{ and }\ x-i$

are two factors of this function.

So, the function of least degree is

$f(x)=\left(x+\dfrac{1}{3}\right)(x+i)(x-i)=\left(x+\dfrac{1}{3}\right)(x^2-i^2)=\\ \\ =\left(x+\dfrac{1}{3}\right)(x^2+1)=\dfrac{1}{3}(3x+1)(x^2+1)=\dfrac{1}{3}(3x^3+x^2+3x+1)$