Find the nth term of this sequences 20,18,16,14

The average heights of four samples taken from a population of students are shown in the table. Which of these is most likely cl

Ne4ueva [31]

Answer:

It would be D, 62

Step-by-step explanation:

The sample size for the group with the average height is the highest, being at 40 instead of 30, 20, or 10, and therefore has the most accurate average height.

7 0

2 years ago

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It takes 8 minutes for Byron to fill the kiddie pool in the backyard using only a handheld hose. When his younger sister is impa

andreyandreev [35.5K]

Let us say that the procedure of completely filling up the pool with water is called “1 job”. Therefore the rates of the equipment in doing the job would be:

Rate of handheld hose = 1 job / 8 minutes ---> 1

Rate of lawn sprinkler = r ---> 2

Rate of the two combines = 1 job/ 5 minutes ---> 3

So we are to find the equation to use, in this case, we simply have to add equations 1 and 2 to get 3:

1 / 8 + r = 1 / 5

Multiplying both sides by 5:

(5 / 8) + 5 r = 1 ---> ANSWER

8 0

2 years ago

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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

1 year ago

A paint manufacturer made a modification to a paint to speed up its drying time. Independent simple random samples of 11 cans of

nydimaria [60]

This question is incomplete. I got the complete part (the boldened part) of it from google as:

The following 98% confidence interval was obtained for μ1 - μ2, the difference between the mean drying time for paint cans of type A and the mean drying time for paint cans of type B:

4.90 hrs < μ1 - μ2 < 17.50 hrs.

Answer:

A paint manufacturer made a modification to a paint to speed up its drying time. Independent simple random samples of 11 cans of type A (the original paint) and 9 cans of type B (the modified paint) were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows. Type A Type B x 1x1equals=76.3 hr x 2x2equals=65.1 hr s 1s1equals=4.5 hr s 2s2equals=5.1 hr n 1n1equals=11 n 2n2equals=9 The following 98% confidence interval was obtained for mu 1μ1minus−mu 2μ2, the difference between the mean drying time for paint cans of type A and the mean drying time for paint cans of type B. What does the confidence interval suggest about the population means?

The mean difference for the 98% confidence interval, the drying times of the two types of paints are (4.90, 17.50). This implies that Type A paint takes between 4.90 and 17.50 hours more to dry than type B paint.

Step-by-step explanation:

The mean difference for the 98% confidence interval, the drying times of the two types of paints are (4.90, 17.50). This implies that Type A paint takes between 4.90 and 17.50 hours more to dry than type B paint.

Only positive values comprise the confidence interval which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification appears to be effective in reducing drying times.

7 0

2 years ago

You are a school photographer taking individual and class pictures for 2 classes of 21 students each. On average, each individua

kramer

146 minutes in total. One student =3 minutes. So then 42x3= 126. Class picture is 10x2=20. So then add both of them you get 146.

3 0

1 year ago

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