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3 years ago

6

Which of the following statements are true about line segments?

1 answer:

Minchanka [31]3 years ago

8 0

Which of the following statements are true about line segments

Answer: that connect two endpoints

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What is the product of the binomials below?

Otrada [13]

Answer:

The correct answer is 8x^2+28x+20.

Step-by-step explanation:

(2x+5)(4x+4)

Multiply the first bracket with each element of second bracket:

=4x(2x+5)+4(2x+5)

=8x^2+20x+8x+20

Now solve the like terms:

=8x^2+28x+20

Thus the correct answer is 8x^2+28x+20....

4 0

4 years ago

What is the value of the discriminant for the quadratic equation 0 = x + 2 + x2? Discriminant = b2 – 4ac

navik [9.2K]

Ax^2 + bx + c = 0
x^2 + x + 2 = 0

a = 1
b = 1
c = 2

b^2 - 4ac

1^2 - 4(1)(2)
= 1 - 8
= -7

The value of the discriminant is -7.

3 0

3 years ago

Read 2 more answers

Camp Ground A has an area of 300 square yards with a maximum capacity of 25 pup tents. Camp Ground B has an area of 480 square y

Damm [24]

Answer:

40 pup tents

Step-by-step explanation:

Since Ground A has area 300 and maximum capacity is 25 tents, so each tent would take up area of 300/25 = 12 sq yards.

Since Ground B's tents should occupy same area as Ground A's tents (12 sq. yards), the maximum capacity of this tent (480 sq yd) would be:

480/12 = 40 pup tents

This is the maximum capacity.

5 0

3 years ago

What is equivalent to five and 300

julia-pushkina [17]

X=120
2/5-(x/300)=0
Simplify= X/300
2/5-X/300=0
Simplify=2/5
2/5-X/300=0
The left denominater is: 5
The right denominater is: 300

6 0

3 years ago

Suppose you know the length of a confidence interval of a population mean is 8.4 and the sample mean (x bar) is 10. Find the ma

SOVA2 [1]

Answer:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The lenght of the interval correspond to:

$8.4 = 2ME$

$ME= \frac{8.4}{2}= 4.2$

And since we know the margin of error we can find the limits for the confidence interval:

$Lower = 10 -4.2=5.8$

$Upper = 10 +4.2=14.2$

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

$\bar X=10$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n represent the sample size

Solution to the problem

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

The sample mean $\bar X$ is distributed on this way:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The margin of error is given by:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The lenght of the interval correspond to:

$8.4 = 2ME$

$ME= \frac{8.4}{2}= 4.2$

And since we know the margin of error we can find the limits for the confidence interval:

$Lower = 10 -4.2=5.8$

$Upper = 10 +4.2=14.2$

4 0

3 years ago