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sladkih [1.3K]
3 years ago
12

The event coordinator asks you to determine how many students participated in the track-and-field day. The total number of stude

nts in seventh and eighth grade combined is 584; 5/8 of them are seventh graders and 3/8 of them are eighth graders. If 4/5 of the seventh graders participated in track-and-field day and 7/8 of the eighth graders participated, about how many total students participated? Describe the process you used to find your answer.
Mathematics
1 answer:
kakasveta [241]3 years ago
8 0
584 is the total number of students in the school.

5/8 are in seventh grade and 3/8 are in eighth grade.

4/5 of the seventh graders participated in the track meet, so there were (4/5 · 5/8) · 584 students in the seventh grade participating in the track meet.

7/8 of the eighth graders participated, so there were (7/8 · 3/8) · 584 students in the eighth grade participating in the track meet.

So, all together, there were

(4/5 · 5/8) · 584 + (7/8 · 3/8) · 584 students from the school in the track meet.

Let's simplify as you asked:

(4/5 · 5/8) · 584 + (7/8 · 3/8) · 584 = [(4/5 · 5/8) + (7/8 · 3/8)] · 584 (distributive property - factoring)

= [20/40 + 21/64] · 584 (multiply fractions)

= (1/2 + 21/64) 584 (reduce the first fraction to lowest terms)

= (32/64 + 21/64) 584 (getting a common denominator)

= (53/64) 584 (combine/add the two fractions)

= 483.625 (multiply together)

All together, there were: 483.625 students in the meet.
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Graph the line with slope -3/5 passing through the point (-2.3)
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4 0
3 years ago
(a) Write 1.04 x 10as an ordinary number.
Natalija [7]

Answer: here are your answers, Add 3.45 X 1010 and 2.75 X 108. The first number is the same as 345 X 108. Note how as the decimal point moves, the exponent changes. Adding them, we get 347.75 X 108 or – less accurately – 3.48 X 1010.

Add 4.00 X 1012 and 7.55 X 1012. The answer is 11.55 X 1012 or 1.16 X 10 13.

Step-by-step explanation: here are your answers, Add 3.45 X 1010 and 2.75 X 108. The first number is the same as 345 X 108. Note how as the decimal point moves, the exponent changes. Adding them, we get 347.75 X 108 or – less accurately – 3.48 X 1010.

Add 4.00 X 1012 and 7.55 X 1012. The answer is 11.55 X 1012 or 1.16 X 10 13.

4 0
3 years ago
A recent survey indicated that the average amount spent for breakfast by business managers was $9.33 with a standard deviation o
olasank [31]

Answer:

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

Replacing the info given we got:

t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3    

Step-by-step explanation:

Information given

\bar X=9.41 represent the sample mean

s=0.24 represent the sample standard deviation

n=81 sample size  

\mu_o =9.33 represent the value to verify

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is higher than 9.33, the system of hypothesis would be:  

Null hypothesis:\mu \leq 9.33  

Alternative hypothesis:\mu > 9.33  

The statistic is given by:

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

Replacing the info given we got:

t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3    

4 0
4 years ago
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