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3 years ago

Find the dimensions of a rectangle whose perimeter is 22 meters and whose area is 28 square meters

1 answer:

8 0

You can solve this by finding all of the factors of 28 and seeing which ones add up to 22 when doubled.

The factors of 28 are as follows:

28 x 1 28(2) + 1(2) = 56 + 2 = 58 ≠ 22
14 x 2 14(2) + 2(2) = 28 + 4 = 32 ≠ 22
7 x 4 7(2) + 4(2) = 14 + 8 = 22

The dimensions of the rectangle are 4 x 7.

You might be interested in

The price of one share of Starbucks declined five dollars per day for four days in a row. How much did the price of one share ch

myrzilka [38]

Answer:

Each share lost $20 in value over 4 days time

Step-by-step explanation:

$5 loss for 4 days

5 x 4 = 20

$20 loss in 4 days.

5 0

3 years ago

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed

algol [13]

Answer:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

$p_v = P(\chi^2_{5} >5.860)=0.32$

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number: 1, 2 , 3 , 4 , 5 ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are given:

$O_{1}=27$ $O_{2}=31$

$O_{3}=42$ $O_{4}=40$

$O_{5}=28$ $O_{6}=32$

The expected values are given by:

$E_{1} =\frac{1}{6}*200=33.33$ $E_{2} =\frac{1}{6}*200=33.33$

$E_{3} =\frac{1}{6}*200=33.33$ $E_{4} =\frac{1}{6}*200=33.33$

$E_{5} =\frac{1}{6}*200=33.33$ $E_{6} =\frac{1}{6}*200=33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

Now we can calculate the degrees of freedom for the statistic given by:

$df=Categories-1=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >5.860)=0.32$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

4 0

4 years ago

Khan Academy Order Rational Numbers<br><br> Please help, I will mark brainliest.

11111nata11111 [884]

Answer:

C= -1/5

B= -0.4

C= -3/7

Step-by-step explanation:

6 0

3 years ago

A box of Georgia peaches has 3 bad and 12 good peaches. (a) If you make a peach cobbler of 12 peaches randomly selected from the

Eddi Din [679]

Answer:

a) 0.21% probability that there are no bad peaches in the peach cobbler.

b) 99.79% probability of having at least 1 bad peach in the peach cobbler

c) 7.91% probability of having exactly 2 bad peaches in the peach cobbler.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the peaches are chosen is not important. So the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

(a) If you make a peach cobbler of 12 peaches randomly selected from the box, what is the probability that there are no bad peaches in the peach cobbler?

Desired outcomes:

12 good peaches, from a set of 12. So

$D = C_{12,12} = \frac{12!}{12!(12 - 12)!} = 1$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{1}{455} = 0.0021$

0.21% probability that there are no bad peaches in the peach cobbler.

(b) What is the probability of having at least 1 bad peach in the peach cobbler?

Either there are no bad peaches, or these is at least 1. The sum of the probabilities of these events is 100%. So

p + 0.21 = 100

p = 99.79

99.79% probability of having at least 1 bad peach in the peach cobbler

(c) What is the probability of having exactly 2 bad peaches in the peach cob- bler?

Desired outcomes:

2 bad peaches, from a set of 3.

One good peach, from a set of 12.

$D = C_{3,2}*C_{12,1} = \frac{3!}{2!(3-2)!}*\frac{12!}{1!(12 - 1)!} = 36$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{36}{455} = 0.0791$

7.91% probability of having exactly 2 bad peaches in the peach cobbler.

3 0

3 years ago

Do you think this Scatter plot is linear or no? I really can't tell.

MrRa [10]

Answer:

I would say no because the dots are all over the place.

Step-by-step explanation:

It keeps going up then down back up and down again?

4 0

3 years ago