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Please answer quickly I will give brainliest- Kelsey has a list of possible functions. Pick one of the g(x) functions below and

Marizza181 [45]

Answer:

1. Behavior

The g(x) is large and positive when x is large and positive;

The g(x) is large and negative when x is large and negative.

2. y-intercept= -18

3. The zeros are: -3, -2 and 3

Step-by-step explanation:

1. End behavior:

To find this you have to know the leading coefficient of the variable with the highest degree, and whether the degree is even or odd.

(x + 3) = x positive, coefficient 1

(x + 2) = x positive, coefficient 1

(x − 3)= x positive, coefficient 1

The coefficient is 1*1*1= 1 = positive

There are 3 x, so it will be x^3= odd

The behavior for odd and positive will be:

g(x) is + ∞ when x=>+∞

g(x) is -∞ when x=>-∞

or

The g(x) is large and positive when x is large and positive;

The g(x) is large and negative when x is large and negative.

2. Y-intercept

The y-intercept is the value of y when the graph touches the y-axis. The y-axis is located at x=0, so to find the y-intercept you need to put x=0 on the graph function. Some graphs can have two y-intercepts but the graph on the question only has one. The calculation will be:

y-intercept = g(0)= (0+3)(0+2)(0-3)

y-intercept = 3*2*-3

y-intercept= -18

3. Zeros

The zeros mean the value of x that will result as g(x) as zero. At this coordinate, the graph will touch the x-axis since g(0) is located on the x-axis. That is why this coordinate also called an x-intercept. The function is expressed as the product of three things. If any of them is 0, then the result will be 0. So we have 3 zeros

$x_{1}$ +3=0

$x_{1}$ =-3

$x_{2}$ +2=0

$x_{2}$ =-2

$x_{3}$ -3=0

$x_{3}$ =3

The zeros are: -3, -2 and 3

8 0

3 years ago

Based on the data in the two-way table, what is the probability that a person weighs 120 pounds, given that he or she consumes 2

sdas [7]

In the question "Based on the data in the two-way table, what is the probability that a person weighs 120 pounds, given that he or she consumes 2,000 to 2,500 calories per day?" The probability of an event, say A given another event, say B is given by n(A and B) / n(B). Thus the probability that a person weighs 120 pounds, given that he or she consumes 2,000 to 2,500 calories per day is given by number of persons that weigh 120 pounds and consume 2,000 to 2,500 calories per day / number of persons that consume 2,000 to 2,500 calories per day. From the table, the number of persons that weigh 120 pounds and consume 2,000 to 2,500 calories per day is 10 while the number of persons that consume 2,000 to 2,500 calories per day is 110. Therefore, the required probability is 10 / 110 = 1 / 11

3 0

3 years ago

Read 2 more answers

The sum of three consecutive even numbers is 114. what is the smallest of the three numbers?

saul85 [17]

Answer:

36

Step-by-step explanation:

Let's let n be any number.

If we multiply n by 2, we get 2n. Since anything multiplied by 2 is even, 2n must be even. Therefore, let 2n be our first number.

So, the consecutive even number will be (2n+2) followed by (2n+4).

We know that their sum must be 114. Therefore, we can write the following equation:

$(2n)+(2n+2)+(2n+4)=114$

Solve for n. Combine like terms on the left:

$6n+6=114$

Subtract 6 from both sides:

$6n=108$

Finally, divide both sides by 6. Therefore, the value of n is:

$n=108/6=18$

Therefore, our first even number is 2n or 2(18) which equates to 36.

And the consecutive even numbers will be 38 and 40.

So, the smallest of them, the first term, is 36.

4 0

3 years ago

Read 2 more answers

The principal at Crest Middle School, which enrolls only sixth-grade students and seventh-grade students, is interested in deter

AlekseyPX

Answer:

a) [ -27.208 , -12.192 ]

b) New procedure is not recommended

Step-by-step explanation:

Solution:-

- It is much more common for a statistical analyst to be interested in the difference between means than in the specific values of the means themselves.

- The principal at Crest Middle School collects data on how much time students at that school spend on homework each night.

- He/She takes a " random " sample of n = 20 from a sixth and seventh grades students from the school population to conduct a statistical analysis.

- The summary of sample mean ( x1 & x2 ) and sample standard deviation ( s1 & s2 ) of the amount of time spent on homework each night (in minutes) for each grade of students is given below:

Mean ( xi ) Standard deviation ( si )

Sixth grade students 27.3 10.8

Seventh grade students 47.0 12.4

- We will first check the normality of sample distributions.

We see that sample are "randomly" selected.
The mean times are independent for each group
The groups are selected independent " sixth " and " seventh" grades.

The means of both groups are conforms to 10% condition of normality.

Hence, we will assume that the samples are normally distributed.

- We are to construct a 95% confidence interval for the difference in means ( u1 and u2 ).

- Under the assumption of normality we have the following assumptions for difference in mean of independent populations:

Population mean of 6th grade ( u1 ) ≈ sample mean of 6th grade ( x1 )

Population mean of 7th grade ( u2 ) ≈ sample mean of 6th grade ( x2 )

Therefore, the difference in population mean has the following mean ( u ^ ):

u^ = u1 - u2 = x1 - x2

u^ = 27.3 - 47.0

u^ = -19.7

- Similarly, we will estimate the standard deviation (Standard Error) for a population ( σ^ ) represented by difference in mean. The appropriate relation for point estimation of standard deviation of difference in means is given below:

σ^ = √ [ ( σ1 ^2 / n1 ) + ( σ2 ^2 / n2 ) ]

Where,

σ1 ^2 : The population variance for sixth grade student.

σ2 ^2 : The population variance for sixth grade student.

n1 = n2 = n : The sample size taken from both populations.

Therefore,

σ^ = √ [ ( 2*σ1 ^2 / n )].

- Here we will assume equal population variances : σ1 ≈ σ2 ≈ σ is "unknown". We can reasonably assume the variation in students in general for the different grade remains somewhat constant owing to other reasons and the same pattern is observed across.

- The estimated standard deviation ( σ^ ) of difference in means is given by:

σ^ =

$s_p*\sqrt{\frac{1}{n_1} + \frac{1}{n_2} } = s_p*\sqrt{\frac{1}{n} + \frac{1}{n} } = s_p*\sqrt{\frac{2}{n}}\\\\\\s_p = \sqrt{\frac{(n_1 - 1 )*s_1^2 + (n_2 - 1 )*s_2^2}{n_1+n_2-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{n+n-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{2n-2} } \\\\s_p = \sqrt{\frac{(20 - 1 )*s_1^2 + (20 - 1 )*s_2^2}{2(20)-2} } \\\\s_p = \sqrt{\frac{19*10.8^2 + 19*12.4^2}{38} } = \sqrt{135.2} \\\\s_p = 11.62755$

σ^ = 11.62755*√2/20

σ^ = 3.67695

- Now we will determine the critical value associated with Confidence interval ( CI ) which is defined by the standard probability of significance level ( α ). Such that:

Significance Level ( α ) = 1 - CI = 1 - 0.95 = 0.05

- The reasonable distribution ( T or Z ) would be determined on the basis of following conditions:

The population variances ( σ1 ≈ σ2 ≈ σ ) are unknown.
The sample sizes ( n1 & n2 ) are < 30.

Hence, the above two conditions specify the use of T distribution critical value. The degree of freedom ( v ) for the given statistics is given by:

v = n1 + n2 - 2 = 2n - 2 = 2*20 - 2

v = 38 degrees of freedom

- The t-critical value is defined by the half of significance level ( α / 2 ) and degree of freedom ( v ) as follows:

t-critical = t_α / 2, v = t_0.025,38 = 2.024

- Then construct the interval for 95% confidence as follows:

[ u^ - t-critical*σ^ , u^ + t-critical*σ^ ]

[ -19.7 - 2.042*3.67695 , -19.7 + 2.042*3.67695 ]

[ -19.7 - 7.5083319 , -19.7 + 7.5083319 ]

[ -27.208 , -12.192 ]

- The principal should be 95% confident that the difference in mean times spent of homework for ALL 6th and 7th grade students in this school (population) lies between: [ -27.208 , -12.192 ]

- The procedure that the matched-pairs confidence interval for the mean difference in time spent on homework prescribes the integration of time across different sample groups.

- If we integrate the times of students of different grades we would have to make further assumptions like:

The intelligence levels of different grade students are same
The aptitude of students from different grades are the same
The efficiency of different grades are the same.

- We have to see that both samples are inherently different and must be treated as separate independent groups. Therefore, the above added assumptions are not justified to be used for the given statistics. The procedure would be more bias; hence, not recommended.

8 0

3 years ago

the ratio of boys to girls in a class is 5:4 there are 36 students in the class how many students are girls

Alina [70]

First add up the total parts in the ratio.
5 + 4 = 9
Now we need to know the value of each part
36 / 9 = 4
The girls has total of 4 parts, therefore we can just multiply the number of parts by the amount in each part
4 x 4 = 16
So there r 16 girls in the class

4 0

3 years ago

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