Can some one help with this one question please

If there are 16 boys and 32 girls in a room, fill out all of the possible ratios of boys to girls that could be made. They alrea

Rzqust [24]

All of possible ratios of boys to girls that could be made is 8:16; 4:8; 2:4; 1:2
Hope it helps, Brainliest if it <3

8 0

2 years ago

Is 3x+2x=3 a function

trasher [3.6K]

Yes this is a function

8 0

3 years ago

Nakita bought items at a grocery store.  She bought 2 boxes of crackers for $3.50 each.  She used a coupon for $0.80 off the p

olga nikolaevna [1]

Answer: $12.65

Step-by-step explanation: sikeeeeeeeeeeee

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

Find the equation of the parabola that has zeros of x = –1 and x = 3 and a y-intercept of (0,–9). Question 1 options: A) y = 3x2

andrey2020 [161]

Answer:

A

Step-by-step explanation:

Given the zeros are x = - 1 and x = 3 then the factors are

(x + 1) and (x - 3) and the parabola is the product of the factors, that is

y = a(x + 1)(x - 3) ← where a is a multiplier

To find a substitute (0, - 9) into the equation

- 9 = a(0 + 1)(0 - 3) = a(1)(- 3) = - 3a ( divide both sides by - 3 )

3 = a, thus

y = 3(x + 1)(x - 3) ← expand the factors using FOIL

= 3(x² - 2x - 3) ← distribute by 3

= 3x² - 6x - 9 → A

7 0

3 years ago