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mars1129 [50]
3 years ago
5

The total points earned in the football game was 42. The Falcons earned 2/3 of that score. How many points did the other team sc

ore?
Mathematics
1 answer:
Elodia [21]3 years ago
4 0

Step-by-step explanation:

other team earned 1/3,

42/1 * 2/3 = 84/3

so 84/3 is 28

then the Falcons = 28

so other team = total score - the Falcons

which is 42 - 28 = 14

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4/5 (0.8)                                 ...

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2 years ago
Ruth’s car traveled 120 miles during the first race.During the second race it traveled 124 miles.If the car averaged 123 miles o
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During the third race, the car traveled 145 miles.
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3 years ago
A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds tha
ollegr [7]

Answer:

99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Step-by-step explanation:

We are given that a high school principal wishes to estimate how well his students are doing in math.

Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

                          P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of students received a passing grade = 77%

           n = sample of tests = 40

           p = population proportion

<em>Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                           level of significance are -2.5758 & 2.5758}  

P(-2.5758 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<u>99% confidence interval for p</u> = [\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

 = [ 0.77-2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } } , 0.77+2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } } ]

 = [0.5986 , 0.9414]

Therefore, 99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Lower bound of interval = 0.5986

Upper bound of interval = 0.9414

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Simplify -3i sqrt -40<br><br> Enter your answer in simplest radical form
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Answer:

Step-by-step explanation:

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Answer:

I think the answer is 43

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