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3 years ago

Which of the following is a random sample? *

Mathematics

1 answer:

Lynna [10]3 years ago

6 0

Answer:

A teacher put the names of students in a hat and pulled out 10.

Step-by-step explanation:

All the other answers had some factor which would cause the answer to lean a certain way. For example, for answer D, Rebecca sent an email, which can be only sent using the internet, to ask how long people are on the internet. All the people that got the email probably use the internet longer than an average person does, making the answer, biased.

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Help , please and thanks

ira [324]

Answer:

891,962,172,409

try that

Step-by-step explanation:

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3 years ago

Kofi has 4 meters of fabric that he wants to cut into 5 equal-sized pieces. What is true about the length of each piece of the f

Romashka-Z-Leto [24]

Answer:

Each piece is 4/5 m long

Step-by-step explanation:

Kofi has 4 meters of fabric

he cuts them into 5 pieces

4meters/ 5 pieces = length of each piece

4/5 meters long

6 0

3 years ago

Read 2 more answers

PLZ HELP WILL MARK BRAINLIEST

Tatiana [17]

$281,978 ....APE.X -_-

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3 years ago

HigherOrder thing name two rays with the same endpoint in the figure below. Do they form an angle? Explain.

Artist 52 [7]

Answer:

where is the figure?

Step-by-step explanation:

4 0

3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers