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VikaD [51]
3 years ago
7

• The perimeter of a rectangular field is 118 meters.

Mathematics
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

101 meter

Step-by-step explanation:

I think

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P=2a+3b+4c solve for c
Pani-rosa [81]
P = 2a + 3b + 4c
4c = P - 2a - 3b
c = (P - 2a - 3b)/4
5 0
3 years ago
Translate the sentence into an equation.
zavuch27 [327]
Answer:
2*(b-6)=8

Explication:
1.divide both sides by 2
b - 6 = 4

2. move the constant to the right and change the sign
b= 4 + 6

3. Calculate
b= 10

7 0
3 years ago
Lisa drew a picture of a boat. She used the scale factor 1 inch : 6 feet
kotykmax [81]

Answer:

42 feet

Step-by-step explanation:

if each inch equals 6 feet, and if her drawing is 7 inches long, then 7×6 would equal the auctual length of the boat, which is 42 feet.

3 0
2 years ago
A swimming coach needs to choose a team for a relay race. The coach must select 4 of 6 available swimmers
Trava [24]

The number of unique ways is given by the number of possible

combination having distinct members.

The number of unique ways there are to arrange 4 of the 6 swimmers are <u>15 ways</u>.

Reasons:

The given parameters are;

The number of swimmers available = 6 swimmers

The number of swimmers the coach must select = 4 swimmers

Required:

The number of unique ways to arrange 4 of the 6 swimmers.

Solution:

The number of possible combination of swimmers is given as follows;

_4C_6 = \dfrac{6!}{4! \times (6 - 4)!}  = 15

Therefore, the coach can select 4 of the 6 available swimmers in <u>15 unique ways</u>

Learn more here:

brainly.com/question/23589217

8 0
2 years ago
A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa
Norma-Jean [14]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

c) Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation  

\bar X=18.3 represent the sample mean

s=3.72 represent the sample standard deviation

n=25 sample size  

\mu_o =15 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b: P-value  and conclusion

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

Part c

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0
2 years ago
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