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Mice21 [21]
3 years ago
14

Please help me! DO NOT SEND LINK!

Mathematics
1 answer:
svet-max [94.6K]3 years ago
6 0

Answer:

<u>So option 4; </u>\frac{35}{100}<u> is the correct option</u>

Step-by-step explanation:

First, we need to convert \frac{2}{10} so it's denominator is 100. To do so, we will multiply the numerator and denominator by 10

\frac{2}{10} *\frac{10}{10} = \frac{20}{100}

Second, we need to add the two fractions to get the current total distance walked. Remember, when adding fractions, the denominator does not change.

\frac{45}{100} + \frac{20}{100} = \frac{65}{100}

Now that we have the total distance walked, we need to subtract that to 1 mile. 1mile=\frac{1}{1} = \frac{10}{10} =\frac{100}{100}

\frac{100}{100} -\frac{65}{100} = \frac{35}{100}

So option 4; \frac{35}{100} is the correct option

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A. 6<br> B. 2<br> C. 0.4<br> D. -6
Strike441 [17]

Answer:

6

Step-by-step explanation:

Hope thats the correct answer.

6 0
3 years ago
V=1/3πh^2 (3r-h) solve for r
stellarik [79]
R=1v+h([pi]h^2) all over 3
5 0
3 years ago
Suppose the time that it takes a certain large bank to approve a home loan is Normally distributed with mean (in days) μ and sta
kolezko [41]

Answer:

(c) H0 should be rejected

Step-by-step explanation:

Null hypothesis (H0): population mean is equal to 5

Alternate hypothesis (Ha): population mean is greater than 5

Z = (sample mean - population mean) ÷ (sd/√n)

sample mean = 5.3, population mean = 5, sd = 1, n = 500

Z = (5.3 - 5) ÷ (1/√500) = 0.3 ÷ 0.045 = 6.67

Using the normal distribution table, for a one tailed test at 0.01 significance level, the critical value is 2.326

Conclusion:

Since 6.67 is greater than 2.326, reject the null hypothesis (H0)

4 0
3 years ago
Use the sample data and confidence level to construct the confidence interval estimate of the population proportion p. n=500, x=
DiKsa [7]

Answer:

0.3581<x<0.4429

Step-by-step explanation:

Using the formula for calculating the confidence interval of the population proportion p expressed as:

Confidence interval = p ± Z * √p(1-p)/n

p is the population proportion = x/n

p = 200/500

p = 0.4

Z is the z-score at 95% CI = 1.96

n is the sample size = 500

Substituting the given parameters into the formula we will have;

Confidence interval = 0.4 ± 1.96 * √p(1-p)/n

Confidence interval = 0.4 ± 1.96 * √0.4(0.6)/500

Confidence interval = 0.4 ± 1.96 * √0.24/500

Confidence interval = 0.4 ± 1.96 * √0.00048

Confidence interval = 0.4 ± 1.96 * 0.0219

Confidence interval = 0.4±0.04294

Confidence interval = (0.3571, 0.4429)

Hence the confidence interval of the population mean is 0.3581<x<0.4429

5 0
3 years ago
Help me pleaseeeeeeeeee!!!!
Gelneren [198K]
Report this clown who put the first answer he’s trying to get your ip.
8 0
3 years ago
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