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AleksandrR [38]
2 years ago
10

4. Which point on the number line represents 0.3?

Mathematics
2 answers:
seropon [69]2 years ago
8 0

Answer: it’s C because it’s the one that’s closer to 3 if u need help just ask

Step-by-step explanation:

Hope this helps

-3

Dafna11 [192]2 years ago
5 0

Answer:

Point C

Step-by-step explanation:

because the number is 0.3 you need to start at zero on the number line and go forward 3 marks. That will lead you to your answer of 0.3

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1/4 is 0.25 as a fraction
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If a man can run p miles in x minutes, how long will it take him to run q miles at the same rate?
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Givenf(x)=6(8−x)+5.<br> (a) What does the notation f (−2) mean? (b) What is the value of f (−2) ?
Nonamiya [84]

Answer:

b. 65

a. The notation f(-2) means that we want to evaluate the value of the equation when x = -2

Step-by-step explanation:

A. The notation f(-2) means that we want to evaluate the value of the equation when x = -2

B. We will simply substitute the value of 2 for x

f(x) = 6(8-x) + 5

f(-2) = 6(8 - (-2)) + 5

f(-2) = 6(8 + 2) + 5

f(-2) = 6(10) + 5 = 60 + 5 = 65

5 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!

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