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Korolek [52]
1 year ago
13

Aaron wants to buy 2 concert tickets that cost $25.50 each. Aaron's grandfather pays him $4.25 an hour to rake leaves. How many

hours does Aaron have to rake leaves to earn the total cost of the tickets?
Mathematics
2 answers:
alexandr402 [8]1 year ago
8 0

You are going to times 25.50 together and that's 51 so then you are going to divide 51.00 by 4.25 and that's 12


So the answer is 12 

ruslelena [56]1 year ago
4 0
Aaron's goal, or his desired total, would be the cost of both tickets together; $51
Let h be the number of hours Aaron has to work to get $51
Our equation is 4.25h = 51
To find h, we must divide both sides of the equation by 4.25
h = 12
Aaron has to work 12 hours to make $51 at $4.25 a hour
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Two different samples will be taken from the same population of test scores where the population mean and standard deviation are
Alenkinab [10]

Answer:

The sample consisting of 64 data values would give a greater precision.

Step-by-step explanation:

The width of a (1 - <em>α</em>)% confidence interval for population mean μ is:

\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}}

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (<em>n</em>).

That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.

Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.

The two sample sizes are:

<em>n</em>₁ = 25

<em>n</em>₂ = 64

The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.

Width for n = 25:

\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{25}}=\frac{1}{5}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]        

Width for n = 64:

\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]

Thus, the sample consisting of 64 data values would give a greater precision

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1 year ago
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Four people—Rob, Sonja, Jack, and Ang—enter their names into a drawing. The winner receives either a t-shirt or a mug, and which
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Answer:

this is the answer 25

Step-by-step explanation:

i guessed

4 0
1 year ago
1) Which of the following is NOT linear?
kupik [55]

Answer:

Step-by-step explanation:

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8 0
2 years ago
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Nadiya determines that there are no fractions equivalent to 2/10 with a denominator greater than 10, but less than 20, that have
frosja888 [35]

Answer:

Yes, Nadiya is correct. By multiplying the numerator and denominator by 2, the first fraction equivalent to 2/10 is 4/20.

Step-by-step explanation:

To find a fraction equivalent to 2/10, we need to multiply (or divide) the numerator and denominator by the same nonzero whole number.

As we need an equivalent fraction with a denominator greater than 10, we will need to multiply and not divide.

The first nonzero whole number we have is 1. If we multiply the numerator and denominator by 1, we get 2/10. Obviously, 2/10 is equivalent to 2/10 but the denominator is not greater than 10, so it doesn't help us.

The next whole number is 2. When we multiply the numerator and denominator by 2, we get 4/20. The denominator is not less than 20.

If we keep going, we will get 6/30, 8/40 and so on.  

Therefore, Nadiya is correct.

The correct answer is the first one.

8 0
2 years ago
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Let Xn be the random variable that equals the number of tails minus the number of heads when n fair coins are flipped. What is t
Firlakuza [10]

Answer:

the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)

Step-by-step explanation:

If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1

therefore

Xn =X1-X2 = X1- (n-X1) = 2X1-n

thus

E(Xn) =∑ (2*X1-n) p(X1) =  2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0

the variance will be

σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) -  n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²

since

σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4

and

σ²(X1) = E(X1²) - [E(X1)]²

n/4 = E(X1²) - (n/2)²

E(X1²) = n(n+1)/4

therefore

σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²

= n - 2n² = n(1-2n)

σ²(Xn) = n(1-2n)

4 0
1 year ago
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