Ask question

Ask question

All categories

3 years ago

11

How can graphics and/or statistics be used to misrepresent data? where have you seen this done?

1 answer:

Anna71 [15]3 years ago

5 0

So basically you need to go up and then along

You might be interested in

3. What is expressed in decimal form?<br>a. 0.58<br>b. 5.8<br>c. 1.6<br>c<br>d. 625

irga5000 [103]

Answer:

0.58

Step-by-step explanation:

because they are the Decimals

3 0

3 years ago

6. The ancient Babylonians were writing fractions in 1800 BCE. But they did not have a concept of zero until about 1489 years la

crimeas [40]

Answer:

hio

Step-by-step explanation:

hi

3 0

3 years ago

After a late night of studying, Ebony decides to grab a latte before class so she can stay awake through her morning lecture. Sh

Rzqust [24]

Answer:

$P(Same\ Bill) = \frac{1}{3}$

$P(Second$

$P(Both\ Even) = \frac{1}{9}$

$Pr(One\ Odd) = \frac{4}{9}$

$P(Sum < 10) = \frac{1}{3}$

Step-by-step explanation:

Given

$Bills: \$1, \$5, \$10$

$Selection = 2\ bills$

The sample space is as follows:

This implies that we construct possible outcome that Ebony selects a bill, returns the bill and then select another.

This means that there are possibilities that the same bill is selected twice.

So, the sample space is as follows:

$S = \{(1,1), (1,5), (1,10), (5,1), (5,5), (5,10), (10,1), (10,5), (10,10)\}$

$n(S) = 9$

Solving (a): $P(Same\ Bill)$

This means that the first and second bill selected are the same.

The outcome of this are:

$Same = \{(1,1),(5,5),(10,10)\}$

$n(Same\ Bill) = 3$

The probability is:

$P(Same\ Bill) = \frac{n(Same\ Bill)}{n(S)}$

$P(Same\ Bill) = \frac{3}{9}$

$P(Same\ Bill) = \frac{1}{3}$

Solving (a): $P(Second < First\ Bill)$

This means that the second bill selected is less than the first.

The outcome of this are:

$Second < First = \{(1,5), (1,10), (5,10)\}$

$n(Second < First) = 3$

The probability is:

$P(Second$

$P(Second$

$P(Second$

Solving (c): $P(Both\ Even)$

This means that the first and the second bill are even

The outcome of this are:

$Both\ Even = \{(10,10)\}$

$n(Both\ Even) = 1$

The probability is:

$P(Both\ Even) = \frac{n(Both\ Even)}{n(S)}$

$P(Both\ Even) = \frac{1}{9}$

Solving (e): $P(Sum < 10)$

This question has missing details.

The correct question is to determine the probability that, the sum of both bills is less than 10

The outcome of this are:

$One\ Odd = \{(1,10), (5,10), (10,1), (10,5)\}$

$n(One\ Odd) = 4$

The probability is:

$Pr(One\ Odd) = \frac{n(One\ Odd)}{n(S)}$

$Pr(One\ Odd) = \frac{4}{9}$

Solving (d): $P(One\ Odd)$

This question has missing details.

The correct question is to determine the probability that, exactly one of the bills is 0dd

The outcome of this are:

$Sum < 10 = \{(1,1), (1,5), (5,1)\}$

$n(Sum < 10) = 3$

The probability is:

$P(Sum < 10) = \frac{n(Sum < 10)}{n(S)}$

$P(Sum < 10) = \frac{3}{9}$

$P(Sum < 10) = \frac{1}{3}$

3 0

3 years ago

ANSWER THE QUESTION FOR BRAINLIEST<br> ANSWER THE QUESTION FOR BRAINLIEST

ElenaW [278]

Answer:

-3/2

Step-by-step explanation:

the constant of variation is commonly known as <em>k .</em>

<em />

y/x= <em>k</em>

<em />

so -3/2 is your answer

3 0

3 years ago

Read 2 more answers

Write and graph an equation in two variables that shows the relationship between the time t and the distance d traveled Moves 2

tiny-mole [99]

Answer:

the distance equation is

$d=\frac{2}{3}\times t$

Step-by-step explanation:

Let's assume

time in hours as 't'

distance traveled as 'd' meters

We are given

Moves 2 meters every 3 hours

So, firstly we find speed

$speed=\frac{2}{3}$

$k=\frac{2}{3}$

we know that

distance = (speed)*(time)

now, we can plug values

$d=k\times t$

now, we can plug value

$d=\frac{2}{3}\times t$

So, the distance equation is

$d=\frac{2}{3}\times t$

Graph:

6 0

3 years ago