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MakcuM [25]
3 years ago
6

CAN SOMEONE PLEASE HELP ME LABEL THE GRAPH SO I CAN PLOT THE POINTS!!

Mathematics
1 answer:
natka813 [3]3 years ago
3 0

Answer:

Era is x and number of wins is y

Edit: I would go up by two's on the y axis and .5 on the x axis

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Can someone help me with this math homework please!
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Answer:

11.2

Step-by-step explanation:

p =\frac{160.16}{14.3}=\frac{160.16*10}{14.3*10}\\\\p = \frac{1601.6}{143}\\\\

p = 11.2

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One of one million people.
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Please help! see attached
nikdorinn [45]

Answer:

  1. The lowest value of the confidence interval is 0.5262 or 52.62%
  2. The highest value of the confidence interval is 0.5538 or 55.38%

Step-by-step explanation:

Here you estimate the proportion of people in the population that said did not have children under 18 living at home.It can also be given as a percentage.

The general expression to apply here is;

C.I=p+-z*\sqrt{\frac{p(1-p)}{n} }

where ;

p=sample proportion

n=sample size

z*=value of z* from the standard normal distribution for 95% confidence level

Given;

n=5000

<u>Find p</u>

From the question 54% of people chosen said they did not have children under 18 living at home

\frac{54}{100} *5000 = 2700\\\\p=\frac{2700}{5000} =0.54

<u>To calculate the 95% confidence interval, follow the steps below;</u>

  • Find the value of z* from the z*-value table

The value of z* from the table is 1.96

  • calculate the sample proportion p

The value of p=0.54 as calculated above

  • Find p(1-p)

0.54(1-0.54)=0.2484

  • Find p(1-p)/n

Divide the value of p(1-p) with the sample size, n

\frac{0.2448}{5000} =0.00004968

  • Find the square-root of p(1-p)/n

=\sqrt{0.00004968} =0.007048

  • Find the margin of error

Here multiply the square-root of p(1-p)/n by the z*

=0.007048*1.96=0.0138

The 95% confidence interval for the lower end value is p-margin of error

=0.54-0.0138=0.5262

The 95% confidence interval for the upper end value is p+margin of error

0.54+0.0138=0.5538

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