Ask question

Ask question

All categories

3 years ago

6

Does anyone know the answer?

1 answer:

s344n2d4d5 [400]3 years ago

3 0

Answer:

d = 9

Step-by-step explanation:

Calculate the slope m using the slope formula and equate to 3

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (4, 6) and (x₂, y₂ ) = (d, 21)

m = $\frac{21-6}{d-4}$ , that is

$\frac{15}{d-4}$ = 3 ← multiply both sides by d - 4

15 = 3(d - 4)

15 = 3d - 12 ( add 12 to both sides )

27 = 3d ( divide both sides by 3 )

9 = d , that is

d = 9

You might be interested in

I need help with this

VashaNatasha [74]

The answer is b. It cannot be c cause the answer would be a negative number. It cannot be a because there would be more than 100 lollipops altogether. And it cannot be d because the answer would be a fraction. There was 29 lollipops altogether. Hope this helps!

3 0

2 years ago

The polynomial below is a perfect square trinomial of the form A^2 - 2AB + B^2

strojnjashka [21]

Answer:

False

Step-by-step explanation:

$9x^2-18x+36\\\\=(3x)^2- 3x \cdot 6+6^2\\\\\text{It is not in the form of}~ A^2-2AB+B^2$

8 0

2 years ago

What is the measure of A and C, A = (12x + 12), b = (15) C =(3x + 18)°

Gnoma [55]

3x + 18 + 12x + 12
15x + 20

6 0

2 years ago

3n at n=11 <br> please help me im suck at maths

Oliga [24]

It just means multiply when they are right next to each other. the answer is 33

6 0

2 years ago

Read 2 more answers

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