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3 years ago

For which values of x is the function

Mathematics

1 answer:

max2010maxim [7]3 years ago

4 0

Answer:

x = - 5, x = - 11

Step-by-step explanation:

Given

y = $\frac{x+4}{(x+5)(x+11)}$

The denominator cannot be zero as this would make y undefined.

Equating the denominator to zero and solving gives the values that x cannot be, that is

(x + 5)(x + 11) = 0

Equate each factor to zero and solve for x

x + 5 = 0 ⇒ x = - 5

x + 11 = 0 ⇒ x = - 11

Thus

x = - 11 and x = - 5 are excluded values.

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Read 2 more answers

A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.

Pachacha [2.7K]

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

The probability that the birth is a male can be written as

$p(m) = 0.52$ (which corresponds to 52%)

While the probability that the birth is a female can be written as

$p(f) = 0.48$ (which corresponds to 48%)

Here we want to calculate the probability that over 2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

$p(mm)=p(m)\cdot p(m)$

And substituting, we find

$p(mm)=0.52\cdot 0.52 = 0.27$

So, 27%.

In this case, we want to find the probability that both children are female, so the probability

$p(ff)$

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

$p(ff)=p(f)\cdot p(f)$