Answer:
<h2>12) There are 534 students who do not lunch in school.</h2>
Step-by-step explanation:
<h3>12. </h3><h3>Simple solution: </h3><h3>890 ÷ 5 = 178 </h3><h3>Therefore: 1/5 = 178 </h3><h3 /><h3>•To find the 2/5 of the students, we have to multiply 178 by 2</h3><h3>178 × 2 = 356</h3><h3>Therefore, 356 students stay to have lunch.</h3><h3 /><h3>•To find the remaining students who do not lunch in school, we have to subtract 356 to 890.</h3><h3>890 - 356 = 534</h3><h3>Therefore, there are 534 students who do not lunch in school.</h3><h3 /><h3 />
Answer:
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Answer:
a)0.6192
b)0.7422
c)0.8904
d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.
Step-by-step explanation:
Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then
z(p)=
where
- Me is the margin of error from the mean
- s is the standard deviation of the population
a.
z(p)=
≈ 0.8764
by looking z-table corresponding p value is 1-0.3808=0.6192
b.
z(p)=
≈ 1.1314
by looking z-table corresponding p value is 1-0.2578=0.7422
c.
z(p)=
≈ 1.6
by looking z-table corresponding p value is 1-0.1096=0.8904
d.
Minimum required sample size for 0.95 probability is
N≥
where
- z is the corresponding z-score in 95% probability (1.96)
- s is the standard deviation (50)
- ME is the margin of error (8)
then N≥
≈150.6
Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.
Answer:
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Step-by-step explanation: