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3 years ago

Sarah earns $12.70 per hour at the car wash. How much will she make if she works 3.5 hours?

Mathematics

2 answers:

ozzi3 years ago

7 0

Answer: 12.70x3.5= 44.5 every hour.. 44.5x3.5=155.75 dollars a hour.

Step-by-step explanation:

uysha [10]3 years ago

7 0

Answer:

$44.45 is what she would earn.

Step-by-step explanation:

Multiply the two numbers together. Batta bing batta boom.

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The U.S Census Brureau has a population clock on the internet.On a recent day the United States population was listed as 310,763

goldfiish [28.3K]

Three hundred ten million, seven hundred sixty three thousand, one hundred thirty six

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2 years ago

Lim x tends to π÷2 , then what is the value of (tanx + cotx) ÷ ( tanx - cotx)

KiRa [710]

(tan(x) + cot(x)) / (tan(x) - cot(x)) = (tan²(x) + 1) / (tan²(x) - 1)

… = (sin²(x) + cos²(x)) / (sin²(x) - cos²(x))

… = -1/cos(2x)

Then as x approaches π/2, the limit is -1/cos(2•π/2) = -sec(π) = 1.

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2 years ago

Find five rational numbers between −2 and 0

Harlamova29_29 [7]

Answer:

-9/5 , -8/5 , -7/5, -6/5 , -5/5.

Step-by-step explanation:

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6 0

3 years ago

Consider the hypothesis test below.

Pavel [41]

Answer:

$Z_{H0} =1.79$

p-value: 0.0367

Decision: Reject H₀

Step-by-step explanation:

Hello!

Hypothesis to test:

H₀:ρ₁-ρ₂=0

H₁:ρ₁-ρ₂>0

The statistic to use to test the difference between two population proportions is the approximation of Z

Z= (^ρ₁-^ρ₂)-(ρ₁-ρ₂) ≈N(0;1)

√ (^ρ₁(1-^ρ₁))/n₁)+(^ρ₂(1-^ρ₂)/n₂))

Z= (0.28-0.15)-0 = 1.79

√ (0.28(1-0.28)/200)+(0.15(1-0.15)/300)

p-value

Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

P(Z>1.79)= 0.0367

Conclusion:

Comparing the p-value against the significance level, you can decide to reject the null hypothesis.

I hope you have a SUPER day!

8 0

3 years ago

Suppose that six individuals are interested in taking part in a study relating BMI to a number of health outcomes. The following

posledela

Answer:

a) Mean = 27.65

Median = 27.645

b) Relative Frequency = 33.33%

Step-by-step explanation:

We are given the following data set:

25.78, 21.06, 36.54, 29.51, 18.96, 34.05

a) Mean and Median

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{165.9}{6} = 27.65$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

Sorted data: 18.96, 21.06, 25.78, 29.51, 34.05, 36.54

$\text{Median} = \displaystyle\frac{25.78 +29.51}{2} = 27.645$

b) BMI above 30 is considered obese

Frequency of obese in the given sample = 2

Relative Frequency =

$\displaystyle\frac{\text{Frequency of obese}}{\text{Total number}} = \frac{2}{6} = 0.3333 = 33.33\%$

5 0

2 years ago