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lidiya [134]
2 years ago
6

The Sydney Morning Herald - February 8, 2004 reported that in 1961, the average number of children born to Australian women (3.5

5) was at its highest level since reliable records began in the 1920s. With only the fact of the mean being 3.55 children, can one make the claim that women were statistically more likely to have four children than any other number of children?
a. Agree, 4 children is the most typical number of children.
b. Agree, 3.55 is closer to 4 than any other whole number.
c. Disagree, the most typical number of children is 3.
d. Disagree, it is impossible to tell
Mathematics
1 answer:
Alborosie2 years ago
3 0

Answer:

<h2>A. Agree, 4 children is the most typical number of children.</h2>

Step-by-step explanation:

By analysis the problem, we can make our  decision by first determining the percent of 3.55 of 4

therefore the percentage can be computed as

=(3.55/4)*100

=0.8875*100

=88.75%

The analysis above it shows that 88.75 percent of the women who give birth to children give birth to 4 children.

This number is very close to a hundred percent hence the statistical claim is  "Agree, 4 children is the most typical number of children."

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Gallons per day.......gallons / day.....so u put the gallons over the number of days, then divide

(1/2) / 6 = 1/2 * 1/6 = 1/12 of a gallon per day <==
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2 years ago
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Delilah and a group of friends had excellent service at a restaurant. They would like to give the server a 25 percent gratuity o
nikklg [1K]

Answer:

$35

Step-by-step explanation:

There is no attached diagram, but despite this, the tip amount can be calculated, since we know the total value of the expense in the restaurant and the percentage of tips they want to give, therefore it would be the multiplication of the total by the percentage like this:

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2 years ago
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A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The
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Answer:

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(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

-1.96 \leq \mu_1 -\mu_2 \leq 7.96

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

\bar X_1= 75.1 represent the sample mean for the scores of the undergraduate students

s_1 = 12.8 represent the standard deviation for the undergraduate students

n_1 =35 the sample size for the undergraduate

\bar X_2= 72.1 represent the sample mean for the scores of the high school students

s_2 = 14.6 represent the standard deviation for the high school students

n_2 =50 the sample size for the high school

The confidence interval for the true difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given by:

df=n_1 +n_2 -2= 35+50-2=83

The confidence level is 90% and the significance level is \alpha=0.1 and \alpha/2 =0.05 then the critical value would be:

t_{\alpha/2}= 1.99

And replacing the info we got:

(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96

(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

-1.96 \leq \mu_1 -\mu_2 \leq 7.96

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

8 0
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The equation r = 6t gives the number of albums released, r, by a record label over time in years, t. If you graph this relations
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R = 6t.....subbing in (8,48).....t = 8 and r = 48
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so u have 2 sets of points on this line and they are (8,48) and (13,78)
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Answer:

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Step-by-step explanation:

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8 0
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