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Luden [163]
3 years ago
6

Order the numbers by value. LEAST to GREATEST10√99√7599.5​

Mathematics
1 answer:
beks73 [17]3 years ago
4 0

Answer:

√75, 9, 9.5, √99, 10

Step-by-step explanation:

that's what I got

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If you’re doing Kaleel to John (not the other way around), you could have 3:1, 3/1, or 3 to 1. Hope this helps!
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The avocado own a 1/4 acre orchard. Two fifths of the orchard  is planted in orange trees. What fraction of an acre in planted i
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Two\ fifths\ of\ the\ orchard= \frac{2}{5}  \cdot  \frac{1}{4}\  acre= \frac{1}{10} \ acre\\\\Ans.\ 0.1\ acre\  in\ orange\ trees.
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Given X(0, 2), Y(-2, 2), and Z(-2, 0), find the coordinates of X', Y', and Z' after a dilation with scale factor -4. ​
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3 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0
3 years ago
This is what I got,
USPshnik [31]
1/x + 1/y = 8
x + y = 8
6/x + 12/y = 72n
4 0
4 years ago
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