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3 years ago

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A farmer needs to enclose a section of land in the shape of a parallelogram by using one side of a barn for one side the barn in

the side opposite of the barn will both be 11 feet long one of the other side will be 8 feet long what is the length of the last side

2 answers:

Pepsi [2]3 years ago

8 0

since 2 sides will be 11 and the third side will be 8 the fourth side also needs to be 8 to make it a parallelogram

Ne4ueva [31]3 years ago

7 0

8 feet long. This is because a parallelogram has same side lengths on both opposite sides. If the first two are 11, the others are 8.

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A study of recent records of car accidents revealed that average proportion of drivers who were distracted by their phones at th

aksik [14]

Answer:

At least 547 records need to be studied.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

And the margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence interval

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $z = 1.96$ .

In this problem, we have that:

$M = 0.04, p = 0.35$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.35*0.65}{n}}$

$0.04\sqrt{n} = 0.93486$

$\sqrt{n} = 23.37$

$n = 546.2275$

At least 547 records need to be studied.

6 0

3 years ago

a math teacher claims that she has developed a review course that increases the score of students on the math portion of a colle

madam [21]

Answer:

A.

$H_0: \mu\leq514\\\\H_1: \mu>514$

B. Z=2.255. P=0.01207.

C. Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.

D. P(z>0.94)=0.1736

Step-by-step explanation:

<em>A. state the null and alternative hypotheses.</em>

The null hypothesis states that the review course has no effect, so the scores are still the same. The alternative hypothesis states that the review course increase the score.

$H_0: \mu\leq514\\\\H_1: \mu>514$

B. test the hypothesis at the a=.10 level of confidence. is a mean math score of 520 significantly higher than 514? find the test statistic, find P-value. is the sample statistic significantly higher?

The test statistic Z can be calculated as

$Z=\frac{M-\mu}{s/\sqrt{N}} =\frac{520-514}{119/\sqrt{2000}}=\frac{6}{2.661}=2.255$

The P-value of z=2.255 is P=0.01207.

The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

Then we can conclude that the score of 520 is significantly higher than 514, in this case, specially because the big sample size.

C. do you think that a mean math score of 520 vs 514 will affect the decision of a school admissions adminstrator? in other words does the increase in the score have any practical significance?

Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.

D. test the hypothesis at the a=.10 level of confidence with n=350 students. assume that the sample mean is still 520 and the sample standard deviation is still 119. is a sample mean of 520 significantly more than 514? find test statistic, find p value, is the sample mean statisically significantly higher? what do you conclude about the impact of large samples on the p-value?

In this case, the z-value is

$Z=\frac{520-514}{s/\sqrt{n}} =\frac{6}{119/\sqrt{350}} =\frac{6}{6.36} =0.94\\\\P(z>0.94)=0.1736>\alpha$

In this case, the P-value is greater than the significance level, so there is no evidence that the review course is increasing the scores.

The sample size gives robustness to the results of the sample. A large sample is more representative of the population than a smaller sample. A little difference, as 520 from 514, but in a big sample leads to a more strong evidence of a change in the mean score.

Large samples give more extreme values of z, so P-values that are smaller and therefore tend to be smaller than the significance level.

8 0

3 years ago

What are 1x3 and 1x3 combined?

goldenfox [79]

Answer:

it would be 6

Step-by-step explanation:

4 0

3 years ago

The personnel files of all eight employees at the Pawnee location of Acme Carpet Cleaners Inc. revealed that during the last six

irina [24]

Answer:

a) For Pawnee:

Range = 11, Mean = 3.13, Mean deviation = 2.44

For Chickpee:

Range = 7, Mean = 4.25, Mean deviation = 1.59

b) Pawnee Location has fewer lost days.

c) variation of the Chickpee is less than Pawnee location

Step-by-step explanation:

Data given:

For Pawnee:

3 1 0 0 2 11 5 3

Max = 11

Min = 0

Range = Max - Min

Range = 11-0 = 11

Range = 11

For Chickpee:

3 6 8 4 4 5 3 1

Max = 8

Min = 1

Range = Max - Min

Range = 8 - 1

Range = 7

For mean of Pawnee:

Mean = (3 + 1 + 0 + 0 + 2 + 11 + 5 + 3)/8

Mean = 25/8

Mean = 3.125

For mean deviation, we find the difference from the mean of each point.

Note: all the negative values will be taken as 0.

Differences from the mean of each point =

(3-3.13)=0.13, (1-3.31) = 2.13 ,

(0-3.13)=0, (0-3.13)=3.13,

(2-3.13)=1.13, (11-3.13)=7.87,

(5-3.13)=1.87, (3-3.13)=0.13

So, our differences are:

(0.13, 2.13, 0, 3.13, 1.13, 7.87, 1.87, 0.13 )

Mean deviation = Sum of all differences/8

Mean deviation = (0.13+2.13+3.13+3.13+ 1.13+ 7.87+ 1.87+ 0.13)/8

Mean deviation = 2.44

For mean of Chickpee:

Mean = (3+6+8+4+4+5+3+1)/8

Mean = 34/8

Mean = 4.25

Similarly, we need to find mean deviation for Chickpee, for that we need to find the differences first as done above for Pawnee.

Note: all the negative values will be taken as 0

Differences from the mean:

(3-4.25) = 1.25, (6-4.25) =1.75,

(8-4.25) = 3.75, (4-4.25)=0.25,

(4-4.25)=0.25, (5-4.25)= 0.75,

(3-4.25)=1.25, (1-4.25) =3.25

So, the differences are:

Differences = (1.25,1.75,3.75,0.25,0.25,0.75,1.25,3.25)

Mean deviation = (1.25+ 1.75+ 3.75+ 0.25+ 0.25+ 0.75+ 1.25+ 3.25)/8

Mean deviation = 1.59

b) which location has fewer lost days:

This can be found out by the total number of days of Pawnee and chickpee.

Pawnee = Sum of values = (3 + 1 + 0 + 0 + 2 + 11 + 5 + 3) = 25 days

Chickpee = Sum of values = (3+6+8+4+4+5+3+1) = 34 days

Hence, Pawnee Location has fewer lost days.

C) which location has less variation?

Formula for variation = ∑( $\frac{x^{2} }{8}$ ) - ( $mean^{2}$ )

where x = values of the set.

For Pawnee:

$mean^{2}$ = $3.13^{2}$ = 9.8

Sum of square of all the data points of Pawnee = 169

Variation = 169/8 - 98 = 11.325

Similarly,

For Chickpee:

$mean^{2}$ = $4.25^{2}$ = 18.06

Sum of square of all the data points of Chickpee = 176

Variation = 176/8 - 18.06

Variation = 22-18.06

Variation = 4

Hence, variation of the Chickpee is less than Pawnee location

8 0

3 years ago

F equals to 2 f - 20

ICE Princess25 [194]

Answer:

20

Step-by-step explanation:

f = 2f - 20

f - 2f = - 20

- f = - 20

f = 20

7 0

3 years ago