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saveliy_v [14]
3 years ago
6

Martha estimated there were 91 marbles in a jar for a contest. The actual number of marbles in the jar was 113. What was the per

cent error of Martha's estimation?
A. 24.18%
B. 80.53%
C. 22%
D. 19.47%
Mathematics
2 answers:
lisabon 2012 [21]3 years ago
5 0

Answer:

D. 19.47

Step-by-step explanation:

91/113 = 80.53

100.00-80.53 = 19.47

babymother [125]3 years ago
4 0

Answer:

19.47% to the nearest hundredth.

Step-by-step explanation:

Value of the error = 113 - 91 = 22.

Percent error = (22/113) * 100

= 19.47.

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Step-by-step explanation:

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-5x + 18 = 3x - 38 <br> show steps please need this asap
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Triangles ABD and ACE are similar right triangles. which ratio best explains why the slope of AB is the same as the slope of AC
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3 years ago
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A simple random sample of size n equals 200 individuals who are currently employed is asked if they work at home at least once p
____ [38]

Answer: 99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

Step-by-step explanation:

<u>step 1:-</u>

Given sample  size n=200

of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week

 Population proportion of employed individuals who work at home at least once per week  P = \frac{x}{n} =\frac{41}{200} =0.205

Q=1-P= 1-0.205 = 0.705

<u>step 2:-</u>

Now  \sqrt{\frac{P Q}{n} } =\sqrt{\frac{(0.205)(0.705)}{200} }

=0.0015

<u>step 3:-</u>

<u>Confidence intervals</u>

<u>using formula</u>

(P  -  Z_∝} \sqrt{\frac{P Q}{n},} (P  +  Z_∝} \sqrt{\frac{P Q}{n},

(0.205-2.58(0.0015),0.205+2.58(0.0015)\\0.20113,0.20887

=0.20113,0.20887[/tex]

<u>conclusion:</u>-

99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

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