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Ber [7]
3 years ago
13

What decimal is the largest? 0.025 0.026 0.25 0.24

Mathematics
2 answers:
riadik2000 [5.3K]3 years ago
8 0
I think the answer is 0.026 because 26 is the biggest number
nasty-shy [4]3 years ago
3 0
0.25 is the largest decimal.
hope i helped;)
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12 with a remainder of 5

Step-by-step explanation:

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kondaur [170]
The correct answer is 3:4
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Andrej [43]

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2 years ago
A window in the shape of a rectangle as shown below has a width of x+5 and a length of x^2- 3x+7 express the area of the rectang
Ksju [112]

Answer:

x^3+x^2-5x+28

Step-by-step explanation:

since a=lw, the area is (x^2-3x+7)(x+4)

1. distribute parentheses \left(x^2-3x+7\right)\left(x+4\right)=x^2x+x^2\cdot \:4+\left(-3x\right)x+\left(-3x\right)\cdot \:4+7x+7\cdot \:4

2. apply +(-a)=-a rule x^2x+4x^2-3xx-3\cdot \:4x+7x+7\cdot \:4

3. Simplify

    Steps to simplify:

x^2x=x^3

3xx=3x^2

3\cdot \:4x=12x

7\cdot \:4=28

x^3+4x^2-3x^2-12x+7x+28

4. Add like terms =x^3+4x^2-3x^2-5x+28

5. Add like terms =x^3+x^2-5x+28

6 0
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
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