HELP PLEASE I WILL MARK BRAINLIEST

Proportional relationships! Need help on!

Jlenok [28]

Direct variation has this equation:

y = kx

where k is the constant of variation

y = -5 ; x = -15

y = kx
-5 = k(-15)
-5/-15 = k
1/3 = k

Choice D. y = 1/3 x

7 0

3 years ago

John’s teacher wants to buy giant cookies for the entire class of cookies cost $2.40 each write an equation that shows how many

krek1111 [17]

40 ÷ 2.40= 16.66
the teacher can buy 16 cookies

4 0

3 years ago

I need help asap. Evaluate Expressions

Anit [1.1K]

Answer:

1:

p-r/6=-6-(-6)/6=-6+1=-5

2:

p-(m-n)= -1-(-4-4)=-1-(-8)=-1+8=7

3:

(z+y)/2)3=(-5-4)/3=-9/3=-3

4:

m/6-n=6/6-6=1-6=-5

5:

k³-h=3³-(-2)=27+2=29

6:

p-(p-(m-3))=5-(5-(4-3))=5-(5-1)=5-4=1

7:

(k)(k-j)+3=(5)(5-5)+3=5*0+3=3

8.

p²/6-q=6²/6-4=6-4=2

9;

zx+3³=6*4+3³=24+27=51.

10:

y+z-(y-x)=5+4-(5-2)=9-3=6

Some basic rules:

+. +. +=+

+. * +=+

+. ÷ +=+

+. - +=+

and

+. +. -=add and put sigh of greater one.

-. * +=-sigh

-÷-=+sigh

- - -=add and put - sigh

7 0

3 years ago

Read 2 more answers

The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co

Alona [7]

Answer:

95% Confidence interval for the variance:

$3.6511\leq \sigma^2\leq 34.5972$

95% Confidence interval for the standard deviation:

$1.9108\leq \sigma \leq 5.8819$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

$s^2=2.89^2=8.3521$

The confidence interval for the variance is:

$\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}$

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

$\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128$

Then, the confidence interval can be calculated as:

$\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972$

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

$\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819$

6 0

3 years ago

Equivalent expressions to 16x-2-24x+4

Tanya [424]

Can i see the Answer Choices?

5 0

3 years ago