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3 years ago

10

A survey was conducted to study the relationship between the annual income of a family and the amount of money the family spends

on entertainment. Data were collected from a random sample of 280 families from a certain metropolitan area. A meaningful graphical display of these data would be:
(a) side-by-side boxplots
(b) a pie chart
(c) a stemplot
(d) a scatterplot
(e) a contingency table

1 answer:

lisov135 [29]3 years ago

6 0

Answer:

The correct option is d i.e. scatter plot

Step-by-step explanation:

The correct option is d i.e. scatter plot

scatter plot will be the best option to display the variation of expenditure with respect to annual income.

on one axis annual income is used and on the other expenditure of the family. hence for a particular change in annual income, an impact on expenditure will easily be predicted.

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3 years ago

Assume the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons.

Rama09 [41]

The rate of change is 16.8 million tons per year

<em><u>Solution:</u></em>

Given that, the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons

<em><u>The average rate of change is given by formula:</u></em>

$\text{ Average rate of change } = \frac{\text{ change in value }}{\text{ number of years }}$

Value in 1950 = 12 million tons

Value in 1955 = 96 million tons

Change in value = value in 1955 - value in 1950

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<em><u>Substitute the given values in formula,</u></em>

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3 years ago

For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure.

Finger [1]

Answer:

Lower limit: 113.28

Upper limit: 126.72

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 120, \sigma = 8$

Middle 60%

So it goes from X when Z has a pvalue of 0.5 - 0.6/2 = 0.2 to X when Z has a pvalue of 0.5 + 0.6/2 = 0.8

Lower limit

X when Z has a pvalue of 0.20. So X when $Z = -0.84$

$Z = \frac{X - \mu}{\sigma}$

$-0.84 = \frac{X - 120}{8}$

$X - 120 = -0.84*8$

$X = 113.28$

Upper limit

X when Z has a pvalue of 0.80. So X when $Z = 0.84$

$Z = \frac{X - \mu}{\sigma}$

$0.84 = \frac{X - 120}{8}$

$X - 120 = 0.84*8$

$X = 126.72$

4 0

3 years ago

An Impressionist painting increases in value

jok3333 [9.3K]

Answer:

A = $8406.6

Step-by-step explanation:

Given:

Average rate $r=9\%$

Initial cost of painting $a = \$1500$

Time $t = 20\ years$

We need to find the final amount of painting at the end of a 20-year.

Solution:

Using Exponential Growth rate formula as:

$A = a(1+r)^t$ ----------(1)

Where:

A = Final amount

a = Initial amount.

r = Rate as a decimal.

t = Time.

Now, we substitute all given values in equation 1.

$A = 1500(1+0.09)^{20}$

$A = 1500(1.09)^{20}$

Substitute $1.09^{20} = 5.60$ in above equation.

$A = 1500\times 5.60$

A = $8406.62

Therefore, value of the painting at the end of a 20-year A = $8406.6

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