Write a trinomial whose binomial factors contain constant terms that sum to −11 and have a product of 18.

The mean time it takes to walk to the bus stop is 8 minutes (with a standard deviation of 2 minutes) and the mean time it takes

kifflom [539]

We have been given for a normal distribution the mean time it takes to walk to the bus stop is 8 minutes with a standard deviation of 2 minutes. And the mean time it takes for the bus to get to school is 20 minutes with a standard deviation of 4 minutes.

(a) Average time that it would take reach school can be obtained by adding the average times.

8+20 = 28 minutes.

(b) Standard deviation of the trip to school can be found as:

$\sigma =\sqrt{2^{2}+4^{2}}=\sqrt{4+16}=\sqrt{20}=4.47$

Therefore, standard deviation of the entire trip is 4.47 minutes.

(c) Let us first find z score corresponding to 30 minutes. $z=\frac{x-\mu }{\sigma }=\frac{30-28}{4.47}=0.447$

We need to find the probability such that $P(x>30)=P(z>0.447)=0.67$

Therefore, the required probability is 0.67.

(d) If average time to walk to school is 10 minutes, then overall average time for the trip will be 10+20 = 30 minutes.

(e) Standard deviation won't change it will remain 4.47

(f) The new probability will be:

$z=\frac{x-\mu }{\sigma }=\frac{30-90}{4.47}=0$

$P(x>30)=P(z>0)=0.5$

Therefore, probability will be 0.50.

6 0

3 years ago

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Help me what is 6x4 and 3x6

Taya2010 [7]

6 x 4= 24

3 x 6= 18

Hope this helps!!

7 0

3 years ago

A survey of 132 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take

ddd [48]

Answer:

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 66/132 = 0.50

Number of samples n = 132

Confidence level = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.50 +/- 1.645√(0.50(1-0.50)/132)

0.50 +/- 1.645√(0.001893939393)

0.50 +/- 0.071589436011

0.50 +/- 0.072

(0.428, 0.572)

The 90% confidence level estimate of the true population proportion of students who responded "yes" is (0.428, 0.572)

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

7 0

3 years ago

A survey results of a middle school shows that 3/20 of the students played football. what percentage of the students play footba

joja [24]

Answer: 15%

Step-by-step explanation:

From the question, we are informed that a survey results of a middle school shows that 3/20 of the students played football.

The percentage of the students that play football in this schools would be calculated as multiplying the fraction by 100. This will be:

= 3/20 × 100

= 3 × 5

= 15%

8 0

3 years ago

Round<br> 38.86 to 1 decimal place

Sonbull [250]

38.9 is the answer. because it is rounded to the nearest tenth.

5 0

3 years ago

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