Ask question

Ask question

All categories

2 years ago

11

the scatter plot shows the population p of a town t years after 20000 witch is he best liner equation to modle the towns populat

ion from 2000 to 2010

2 answers:

attashe74 [19]2 years ago

6 0

I think it would be c hope this helps you out

nevsk [136]2 years ago

6 0

Answer:

C

Had the same question *we love cheating on IREADY*

You might be interested in

Quadrilateral RSTQ is a parallelogram.

Basile [38]

<S ~= <Q AND TQ ~= SR

3 0

3 years ago

Colin and Brian scored 59. Brian scored 59 more points than Colin ,so what are Colin and brain scored combine together ? Colin s

makvit [3.9K]

Answer:

59

Step-by-step explanation:

Let c and b represent the scores of Colin and Brian respectively. Then

c + b = 59. Since brian scored 59 more points than Colin, that means c = 0 and b = 59. Their combined score is 0 + 59 = 59.

6 0

3 years ago

I really need help would please like help thanks

lawyer [7]

The reciprocal of a fraction $\frac{a}{b}$ is the same fraction "upside down" $\frac{b}{a}$ .

The reciprocal of a whole number $N$ is "1 over" that number $\frac{1}{N}$ .

10/9
11/1 or 11
1/10

8 0

3 years ago

Please Help me!!!<br> Compound Inequalities!!!

Rama09 [41]

Answer is B

9/4 <x <= 4
...........

6 0

3 years ago

To find the standard deviation of the diameter of wooden dowels, the manufacturer measures 19 randomly selected dowels and find

loris [4]

Answer:

Option D - $[0.12$

Step-by-step explanation:

Given : The manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be s=0.16.

To find : The 95% confidence interval for the population standard deviation sigma?

Solution :

Number of sample n=19

The degree of freedom is Df=n-1=19-1=18

The standard deviation of the sample is s=0.16

Applying chi-square table to find critical value,

Upper critical value of $\chi^2$ is $UC=\chi(\frac{0.05}{2},18) = 31.5264$

Lower critical value of $\chi^2$ is

$LC=\chi(1-\frac{0.05}{2},18) = 8.2307$

Lower limit of the 95% confidence interval for the population variance

$L=\frac{(df)\times (s^2)}{UC}$

$L=\frac{18\times (0.16^2)}{31.5264}$

$L=\frac{18\times0.0256}{31.5264}$

$L=\frac{0.4608}{31.5264}$

$L=0.0146$

Upper limit of the 95% confidence interval for the population variance

$U=\frac{(df)\times(s^2)}{LC}$

$U=\frac{18\times (0.16^2)}{8.2307}$

$U=\frac{18\times0.0256}{8.2307}$

$U=\frac{0.4608}{8.2307}$

$U=0.0559$

So, The 95% confidence interval for the population variance is [0.0146, 0.0560]

Now, The 95% confidence interval for the population standard deviation is

$[\sqrt{0.0146}$

$[0.1208$

or $[0.12$

Therefore, Option D is correct.

The 95% confidence interval for the population standard deviation is $[0.12$

6 0

2 years ago