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krok68 [10]
2 years ago
12

What is 0.476 as a fraction?

Mathematics
1 answer:
shutvik [7]2 years ago
8 0

Answer:

\frac{5}{7} *\frac{2}{3} \\\frac{10}{21}

Step-by-step explanation:

Hence 10/21 is the fraction of 0.47619047619

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Step-by-step explanation:

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2 years ago
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Alik [6]

Answer:

One Solution

Step-by-step explanation:

x + 3y = 3      

x - y = -3        

4y = 6           Subtract to solve by elimination.

y = \frac{3}{2}              Divide by 4.

x - \frac{3}{2}  = -3      Substitute \frac{3}{2} for y in the original equation.

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6 0
2 years ago
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

= √{[Σ(x - xbar)²]/(N-1)} = 16.116244392951 = 16.116244 to 6 d.p.

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!!!

4 0
3 years ago
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zlopas [31]

Answer:

Step-by-step explanation:

It maybe will be \neq x^{2} \leq \\ \\ \int\limits^a_b {x} \, dx \int\limits^a_b {x} \, dx \sqrt{x} \\ \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. x^{2} x^{2} \sqrt{x}  \lim_{n \to \infty} a_n  \lim_{n \to \infty} a_n \neq \sqrt{x} \sqrt[n]{x} \frac{x}{y} \frac{x}{y} \alpha \beta x_{123} \\ x^{2} \int\limits^a_b {x} \, dx x^{2}

7 0
3 years ago
Read 2 more answers
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