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3 years ago

What is the length of segment AC? On a coordinate plane, line A C has points (3, negative 1) and (negative 5, 5).

Mathematics

2 answers:

Debora [2.8K]3 years ago

7 0

Answer:

Step-by-step explanation:

Just took the test

VMariaS [17]3 years ago

4 0

the answer is 10 on edge

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Josiah invests $360 into an account that accrues 3% interest annually. Assuming no deposits or withdrawals are made, which equat

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3 years ago

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3 years ago

What is the slope of a line that goes through the points (1, -5) and (-3, 2)?

guajiro [1.7K]

Answer:

Step-by-step explanation:

y2-y1 / x2-x1

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3 years ago

For which values of a the system has at least one solution: x≤7 x≥a urgent pls help now

allochka39001 [22]

Answer:

We have the system:

x ≤ 7

x ≥ a

Now we want to find the possible values of a such that the system has, at least, one solution.

First, we should look at the value of a where the system has only one solution:

We can write the 2 sets as:

a ≤ x

x ≥ 7

So, writing both together:

a ≤ x ≤ 7

if a is larger than 7, we do not have solutions.

then a = 7 gives:

7 ≤ x ≤ 7

Here the only solution is 7.

Now, if a is smaller than 7, for example 5, we have:

5 ≤ x ≤ 7

Now x can take different values, so we have a lot of solutions.

Then the restrictions for a, such that the system has at least one solution, is:

a ≤ 7.

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

3 years ago