All categories

3 years ago

Which type of statistical graphic uses bars to describe the data?

2 answers:

6 0

Histogram. A histogram in another kind of graph that uses bars in its display. This type of graph is used with quantitative data. Ranges of values, called classes, are listed at the bottom, and the classes with greater frequencies have taller bars.

raketka [301]3 years ago

5 0

Answer:

A histogram

Step-by-step explanation:

This type of graph is used with quantitative data. Ranges of values, called classes, are listed at the bottom, and the classes with greater frequencies have taller bars.

You might be interested in

C=48y+0.99n solve for y

Ne4ueva [31]

Answer:

y = (C -0.99n)/48

Step-by-step explanation:

Undo what is done to y, in the reverse order. We have y multiplied by 48 then added to 0.99n. So, we must add the opposite of 0.99n and multiply that result by the reciprocal of 48.

C = 48y + 0.99n . . . . . . given

C - 0.99n = 48y . . . . . . add -0.99n

(C -0.99n)/48 = y . . . . . multiply by 1/48

4 0

3 years ago

Willy says the slope of the graph is -1/3.what error did she make when finding the slope

Nadusha1986 [10]

Answer:

you did not add a picture or any other over fractions.

Step-by-step explanation:

3 0

3 years ago

The interior in a triangle have the following measures: (x+34), (6x+4),and (4x+10). Find the measure of the smallest angle in th

guapka [62]

(6x+4) hope this helped
please leave a thank you

3 0

3 years ago

A study revealed a strong, linear, negative association between a decrease in salary and amount spent on clothes. Another study

kakasveta [241]

Answer:

A. simpson's paradox

Step-by-step explanation:

The Simpson's paradox was named after Edward Simpson, the person who described this paradox for the first time in 1951. In this paradox, you find two contrary patterns. For example, a positive and a negative correlation, depending on how data is analyzed. The differences in the analyses are how data are grouped. This paradox is observed often in social researches. Most of the times, results are affected by the sample on each group or additional information related to the data.

5 0

3 years ago

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0

2 years ago