All categories

3 years ago

Factor: 11x^2+35x+6 I dont know how to do this one either. Lol.

Mathematics

2 answers:

Allushta [10]3 years ago

4 0

The answer is x= -3, or <span> x = -2/11 = -0.182</span>

Flauer [41]3 years ago

4 0

Answer:

(11x+2) (x+3) .

Step-by-step explanation:

Given: 11x²+35x+6 .

To find: Factor

Solution: We have given that

11x²+35x+6

On factoring

11x²+33x+ 2x+6

Taking common 11x from first two term and 2 from last two term

11x(x+3) +2 (x+3)

On grouping

(11x+2) (x+3) .

Therefore, (11x+2) (x+3) .

You might be interested in

N is the set of odd natural numbers greater than 11 and less than 21. Use set notation to write the elements of N. N

S_A_V [24]

Answer:

N = {13, 15, 17, 19}

Step-by-step explanation:

The natural numbers greater than 11 and less than 21 are:

12, 13, 14, 15, 16, 17, 18, 19 and 20

Odd numbers are numbers that have remainders when divided by 2. Out of the numbers above, the numbers with remainders when divided by 2 are 13, 15, 17 and 19.

Since N is the set of odd natural numbers greater than 11 and less than 21,

N = {13, 15, 17, 19}

4 0

3 years ago

Read 2 more answers

9. Simplify (3a4 – 2a2 + 5a – 10) – (2a4 + 4a2 + 5a – 2).

Zinaida [17]

The answer to the question is a

4 0

3 years ago

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

8 0

3 years ago

Do you know what the answer is to this question

RUDIKE [14]

Use subsiution]
y=-3x+5
subsitute -3x+5 for y in second equation
5x-4(-3x+5)=-3
distribute
5x+12x-20=-3
add like terms
17x-20=-3
add 20 to both sides
17x=17
divide both sides by 17
x=1
now subsitute 1 for x
y=-3x+5
y=-3(1)+5
y=-3+5
y=2

x=1
y=2

4 0

3 years ago

Can somebody help me with this please

dmitriy555 [2]

Uhhh, kind of an estimated guess but I got $7426.30

4 0

3 years ago