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3 years ago

Explain how showing fractions with models and a number line are alike and different?

1 answer:

5 0

Showing fractions can be expressed in model and number line.
Using a model, you have to draw the number of parts (denominator) and shade the number of parts (numerator).

Using a number line, the line starts from 0 to the number of the denominator. You put a marker to the number that is the same as the numerator.

Both model and number line use the value of the numerator and denominator. What makes them different is how they are presented. Model can easily be understood compared to number line.

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A sports news station wanted to know whether people who live in the North or the South are bigger sports fans. For its study, 17

Free_Kalibri [48]

Using the z-distribution, it is found that the 95% confidence interval for the difference is (-1.3, -0.7).

<h3>What are the mean and the standard error for each sample?</h3>

Considering the data given:

$\mu_S = 3.8, n = 175, s_S = \frac{1.7}{\sqrt{175}} = 0.1285$

$\mu_N = 4.8, n = 152, s_N = \frac{1.2}{\sqrt{152}} = 0.0973$

<h3>What is the mean and the standard error for the distribution of differences?</h3>

The mean is the subtraction of the means, hence:

$\overline{x} = \mu_S - \mu_N = 3.8 - 4.8 = -1$

The standard error is the square root of the sum of the variances of each sample, hence:

$s = \sqrt{s_S^2 + s_N^2} = \sqrt{0.1285^2 + 0.0973^2} = 0.1612$

<h3>What is the confidence interval?</h3>

It is given by:

$\overline{x} \pm zs$

We have a 95% confidence interval, hence the critical value is of z = 1.96.

Then, the bounds of the interval are given as follows:

$\overline{x} - zs = -1 - 1.96(0.1612) = -1.3$

$\overline{x} + zs = -1 + 1.96(0.1612) = -0.7$

More can be learned about the z-distribution at brainly.com/question/25890103

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4 0

2 years ago

What does b= in mathematics

ziro4ka [17]

bisector. Base. In geometry, the base of a shape is the side (usually the bottom) that forms a right (90 degree) angle with the height of the object. Biannual.

7 0

3 years ago

Identify whether the series sigma notation infinity i=1 15(4)^i-1 is a convergent or divergent geometric series and find the sum

const2013 [10]

Answer: The correct option is

(d) This is a divergent geometric series. The sum cannot be found.

Step-by-step explanation: The given infinite geometric series is

$S=\sum_{i=1}^{\infty}15(4)^{i-1}.$

We are to identify whether the given geometric series is convergent or divergent. If convergent, we are to find the sum of the series.

We have the D' Alembert's ratio test, states as follows:

Let, $\sum_{i=1}^{\infty}a_i$ is an infinite series, with complex coefficients $a_i$ and we consider the following limit:

$L=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}.$

Then, the series will be convergent if L < 1 and divergent if L > 1.

For the given series, we have

$a_i=15(4)^{i-1},\\\\a_{i+1}=15(4)^i.$

So, the limit is given by

$L\\\\\\=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i-1}}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i}4^{-1}}\\\\\\=\dfrac{1}{4^{-1}}\\\\=4>1.$