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pickupchik [31]

3 years ago

8

Find the slope of the line (4,4) (-4,0)

2 answers:

slava [35]3 years ago

8 0

The slope is 0.5, hope this helps

MakcuM [25]3 years ago

6 0

Answer:

your answer should be m=1/2

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Please help? I'll fail if I miss this....

Svetach [21]

The answer is Triangle PQR

6 0

3 years ago

Read 2 more answers

If the numbers -2.5, -1.6, -3 1/10, - 1/10, and -0.5 were graphed on a number line, which would appear farthest to the left?

Leona [35]

Turn the question around.
If all the numbers were positive: 2.5, 1.6, 3 1/10, 1/10, and 0.5, which would be farthest to the right?

The fact is that the negative numbers are a reflection of the positive numbers.
Since 3 1/10 would be farthest to the right with positive numbers, -3 1/10 will be farthest to the left with the negative numbers.

7 0

3 years ago

1. A square and a rectangle have the same area. The length of the rectangle is 32

inessss [21]

Answer:

8

Step-by-step explanation:

A=l*w

32*2=64

A=s*s

s=√A

√64

=8

6 0

3 years ago

Solve and check: -9k+13+k=61

VARVARA [1.3K]

-8k= 48

divide 8

ANSWER: k = -6

8 0

3 years ago

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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