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Andru [333]
2 years ago
12

36 people attended Quintero’s Super Bowl party. Each person cheered for either the Patriots or the Rams. If 15 people were rooti

ng for the Patriots, which of the following equations could be used to find how many , f , we’re cheering for the Rams.
Answers :
A : f - 15 = 36

B : 15 + f = 36

C : f + 36 = 15

D : f = 36 - 15

E : 15f = 36

Thank you helping me whoever answers this

And you’ll get brainliest plus extra points 100
Mathematics
1 answer:
Gnesinka [82]2 years ago
3 0

Answer:

the answer is D. 36 fans in total minus 15 fans that rooted for the rams. Equals the amount of fans that rooted for the Patriots

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Troyanec [42]

Answer:

The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

Step-by-step explanation:

1). Height of the trees at Yard Works are = 7,9,7,12,5 feet

So mean height of the trees = (7+9+7+12+5)÷5

                                               = 40÷5 =8 feet

Standard deviation of the trees at Yard works = ∑(║(height of the tree-mean height of the tree))║/(number of trees)

(height of the tree-mean height of the tree)= ║(7-8)║+║(9-8)║+║(7-8)║+║(12-8)║+║(5-8)║ = (1)+1+(1)+4+(3)= 10

Therefore standard deviation = (10)/(5) =2

2). In the same way mean height of the trees at Grow Station=(9+11+6+12+7)/5= 45/5 = 9

Now we will calculate the mean deviation of the tress at Grow Station

= ∑║(height of the tree-mean height of the tree)║/(number of trees)

= ║(9-9)║+║(11-9)║+║(6-9)║+║(12-9)║+║(7-9)║/(5)

= (0+2+3+3+2)/5

= 10/5 =2

Therefore The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

                                                             

7 0
3 years ago
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Please help me. this might be easy for some and after 2 answers I will give brainliest
Evgen [1.6K]

Answer:

7 yards and 2 feet is 86 feet long.

Step-by-step explanation:

7·12=84

84+2=86

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2 years ago
What is #3? I need to know. ​
MissTica

Answer huh?

Step-by-step explanation:

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2 years ago
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A.) find the test statistic
Arada [10]

Regarding the hypothesis tested in this problem, it is found that:

a) The test statistic is: t = 0.49.

b) The p-value is: 0.6297..

c) The null hypothesis is not rejected, as the p-value is greater than the significance level.

d) It appears that the students are legitimately good at estimating one minute, as the sample does not give enough evidence to reject the null hypothesis.

<h3>What is the test statistic?</h3>

The test statistic of the t-distribution is given by the equation presented as follows:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

In which the variables of the equation are given as follows:

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  • \mu is the value tested at the hypotheses.
  • s is the standard deviation of the sample.
  • n is the sample size.

The value tested at the hypothesis is:

\mu = 60

From the sample, using a calculator, the other parameters are given as follows:

\overline{x} = 62.47, s = 19.2, n = 15

Hence the test statistic is obtained as follows:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{62.47 - 60}{\frac{19.2}{\sqrt{15}}}

t = 0.49.

<h3>What are the p-value and the conclusion?</h3>

The p-value is obtained using a t-distribution calculator, with a two-tailed test, as we are testing if the mean is different a value, in this case 60, with:

  • 15 - 1 = 14 df, as the number of degrees of freedom is one less than the sample size.
  • t = 0.49, as obtained above.

Hence the p-value is of 0.6297.

Since the p-value is greater than the significance level of 0.01, the null hypothesis is not rejected, attesting to the ability of the students to estimate one minute.

More can be learned about the test of an hypothesis at brainly.com/question/13873630

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Step-by-step explanation:

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