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2 years ago

What is the simplified form of the quantity y−squared minus y minus 12 over the quantity y−squared minus 2y minus 15?

1 answer:

6 0

(y^2-y-12)/(y^2-2y-15) factor both the numerator and denominator...

(y^2-4y+3y-12)/(y^2-5y+3y-15)

(y(y-4)+3(y-4))/(y(y-5)+3(y-5))

((y-4)(y+3))/((y-5)(y+3)) so the (y+3) terms cancel leaving

(y-4)/(y-5)

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Find x. select one : a. 3/4 b. 1/2 c. 2 d. 4

Dimas [21]

check the picture below.

8 0

3 years ago

HELPPPPPPP!!!! 0÷0 Find the RATIO and the EXACT VALUE of the given CscB.

AlekseyPX

Answer:

Step-by-step explanation:

take angle B as reference angle

using cosec rule

cosec B=hypotenuse/opposite

cosec B=13/12

cosec B=1.08

B=cosec 67

B=67 degree

for ratio,

cosec B=13/12

7 0

3 years ago

Question 4 options: Find the mean and standard deviation for a binomial distribution with 680 trials and a probability of succes

lys-0071 [83]

Answer:

Mean for a binomial distribution = 374

Standard deviation for a binomial distribution = 12.97

Step-by-step explanation:

We are given a binomial distribution with 680 trials and a probability of success of 0.55.

The above situation can be represented through Binomial distribution;

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

where, n = number of trials (samples) taken = 680 trials

r = number of success

p = probability of success which in our question is 0.55

So, it means X ~ $Binom(n=680, p=0.55)$

Now, we have to find the mean and standard deviation of the given binomial distribution.

Mean of Binomial Distribution is given by;

E(X) = n $\times$ p

So, E(X) = 680 $\times$ 0.55 = 374

Standard deviation of Binomial Distribution is given by;

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

= $\sqrt{680 \times 0.55 \times (1-0.55)}$

= $\sqrt{680 \times 0.55 \times 0.45}$ = 12.97

Therefore, Mean and standard deviation for binomial distribution is 374 and 12.97 respectively.

3 0

3 years ago

For f(x)=1/x^2-3, find a) f(3) b) f(2+h) Show/explain work

valentina_108 [34]

$f(x)=\dfrac{1}{x^2-3}\\\\a)\ f(3)\to\text{put x=3 to the equation of the function f}\\\\f(3)=\dfrac{1}{3^2-3}=\dfrac{1}{9-3}=\dfrac{1}{6}\\\\Answer:\ \boxed{f(3)=\dfrac{1}{6}}\\\\b)\ f(2+h)\to\text{put x=2+h to the equation of a function f}\\\\f(2+h)=\dfrac{1}{(2+h)^2-3}\ \ \ |\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\=\dfrac{1}{2^2+2(2)(h)+h^2-3}=\dfrac{1}{4+4h+h^2-3}=\dfrac{1}{h^2+4h+1}\\\\Answer:\ \boxed{f(2+h)=\dfrac{1}{h^2+4h+1}}$

6 0

3 years ago

Piece-wise defined function been having trouble :,)

LekaFEV [45]

Answer:

c.y is the correct answer

Step-by-step explanation:

because its defined in x from the left ×<=1

3 0

2 years ago