Ask question

Ask question

All categories

2 years ago

11

Why is a scatterplot an appropriate display for this data set?

2 answers:

Nookie1986 [14]2 years ago

5 0

Answer:

It could be because it shows trends and shows outliers and all the data in a way that makes sense

Step-by-step explanation:

sp2606 [1]2 years ago

4 0

Answer:

A scatter plot can also be useful for identifying other patterns in data. We can divide data points into groups based on how closely sets of points cluster together. Scatter plots can also show if there are any unexpected gaps in the data and if there are any outlier points

pls mark as brainliest answer

Step-by-step explanation:

You might be interested in

Please help me i need to turn this in rn

Cloud [144]

Answer:

Step-by-step explanation:

2x6yz

7 0

3 years ago

If "-3(y-10=9" what is the value of y-12

zvonat [6]

Answer:

the correct answer is -5 since y=7

8 0

2 years ago

Read 2 more answers

EASY 7TH GRADE MATH, WILL GIVE BRAINLIEST AND THANKS

Fudgin [204]

Answer:

D. 8

Step-by-step explanation:

2p=16

divide both sides by 2 to get p alone

p=8

7 0

2 years ago

Read 2 more answers

A right triangle has a height of 20 cm and a base of 10 cm. What is the area of the triangle? (A = 1/2 bh)

alisha [4.7K]

Answer:

100 because its way too ez

7 0

2 years ago

Read 2 more answers

A major department store chain is interested in estimating the mean amount its credit card customers spent on their first visit

sergiy2304 [10]

Answer:

95% Confidence interval: (39.43, 61.58)

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = $50.50

Sample size, n = 15

Alpha, α = 0.05

Sample standard deviation = 20

95% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.05} = \pm 2.1447$

$=50.50 \pm 2.1447(\dfrac{20}{\sqrt{15}} ) \\\\= 50.50 \pm 11.0751 \\= (39.4249,61.5751)\\\approx (39.43, 61.58)$

95% Confidence interval: (39.43, 61.58)

8 0

2 years ago