All categories

3 years ago

Finding the slope intercept. 1x+y=-6

Mathematics

1 answer:

joja [24]3 years ago

4 0

Here ya go! So let me walk you through it: first you need to get 1 to the other side. I said -1 because I placed it to the other side; since y=mx+b. So then you put in the slope intercept form. Btw sorry about the text lol..i wrote this in rough

You might be interested in

How do you solve 111.8 divided by 13

rodikova [14]

Line the decimal first then divide 3

4 0

3 years ago

For the graphed exponential equation, calculate the average rate of change from x = 0 to x = 1.

777dan777 [17]

The average rate of change is 4x

6 0

3 years ago

Read 2 more answers

Approximately how many gallons are in

Nataliya [291]

Answer:

1.8

Step-by-step explanation:

cuz

if 1L is equal to 0.26

then 7×0.26

= 1.8

hope it helps

8 0

3 years ago

Read 2 more answers

Please help me with these 2 question.

maks197457 [2]

Answer:

Step-by-step explanation:

- sqrt(39) and square root 47

are the limits. The - square root of 39 is smaller than - 6. So the integer to use here is - 6

sqrt (47) = 6.686. Here the square root is larger than the closest integer.

The integer to use is 6

- 6 + 6 = 0

4 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

3 years ago