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anygoal [31]
3 years ago
12

Ms. Geetha said that 60 of her total students will compete in the Math Olympics. If 60 is 20% of her students, how many total st

udents does she have?
Mathematics
1 answer:
WARRIOR [948]3 years ago
4 0

Answer:

300 students

Step-by-step explanation:

There are two ways to solve this question.

  • Way Number 1

60 is 20% of the total students

20÷100= 0.2

60÷0.2=300

Therefore there are 300 students

  • Way Number 2

60 is 20% of the total students

60/20 = x/100

x= 60*100÷20

x= 6000÷20

x= 300

Therefore there are 300 students.

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Step-by-step explanation:

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A state legislator wants to determine whether his voters' performance rating (0 - 100) has changed from last year to this year.
ra1l [238]

Answer:

Step 1 of 4

Point estimate for the population mean of the paired differences = -8.2

Step 2 of 4

Sample standard deviation of the paired differences = 16.116244

Step 3 of 4

Margin of Error = ±9.326419

Step 4 of 4

90% Confidence interval = (-17.5, 1.1)

Step-by-step explanation:

The ratings from last year and this year are given in table as

Rating (last year) | x1 | 87 67 68 75 59 60 50 41 75 72

Rating (this year) | x2| 85 52 51 53 50 52 80 44 48 57

Difference | x2 - x1 | -2 -15 -17 -22 -9 -8 30 3 -27 -15

Step 1 of 4

Mean = (Σx)/N = (-82/10) = -8.2 to 1 d.p.

Step 2 of 4

Standard deviation for the sample

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Step 3 of 4

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = -8.2

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 10 - 1 = 9.

Significance level for 90% confidence interval

= (100% - 90%)/2 = 5% = 0.05

t (0.05, 9) = 1.83 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 16.116244

n = sample size = 10

σₓ = (16.116244/√10) = 5.0964038367

Margin of Error = (Critical value) × (standard Error of the mean) = 1.83 × 5.0964038367 = 9.3264190212 = 9.326419 to 6 d.p.

Step 4 of 4

90% Confidence Interval = (Sample mean) ± (Margin of Error)

CI = -8.2 ± (9.326419)

90% CI = (-17.5264190212, 1.1264190212)

90% Confidence interval = (-17.5, 1.1)

Hope this Helps!!!

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