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3 years ago

THe chart for my last post

Mathematics

1 answer:

nydimaria [60]3 years ago

3 0

Answer:

Median : 23

Mode: 11 and 24

Range: 28

Step-by-step explanation:

A stem-and- leaf plot is a very special table where each " value" is split into a stem.

Basically the stem is the foirst value of the number and the leaves are the last digits.

<u><em>Example:</em></u> A stem of 1 and a leaf of 1 has a value of 11.

A stem of 1 and a leaf of 2 has a value of 12.

In this case the numbers in the plot are:

11, 11, 12, 13, 14, 15, 22, 23, 24, 24, 27, 32, 35, 37, 39

The questions are :

<h3><u><em>1.</em></u> What is the median? </h3><h3><u><em>2.</em></u> What is the mode ?</h3><h3><u><em>3. </em></u>What is the range ?</h3>

To answer the first question, you need to list them in order.

Then find the center of those numbers. Cross out one number from the left and the right and keep doing it until there is only one ot two numbers.

<u><em>One number- that is the median</em></u>

<u><em>two numbers left - find the mean of the two numbers and that is the median</em></u>

So in this case the median is 23 crossing out seven numbers on each side.

To answer the second question, you need to find the number that comes up the most. In this case, 11 and 24.

To answer the third and last question, you need to find the mininum value and the maximum and find the difference.

In this case, 11 and 39, the difference is 28.

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2 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height

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Answer:

a) h = 0.1: $\bar v = -11\,\frac{ft}{s}$ , h = 0.01: $\bar v = -10.1\,\frac{ft}{s}$ , h = 0.001: $\bar v = -10\,\frac{ft}{s}$ , b) The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

Step-by-step explanation:

a) We know that $y = 30\cdot t -10\cdot t^{2}$ describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ( $\bar v$ ), measured in feet per second, can be done by means of the following definition:

$\bar v = \frac{y(2+h)-y(2)}{h}$

Where:

$y(2)$ - Position of the ball evaluated at $t = 2\,s$ , measured in feet.

$y(2+h)$ - Position of the ball evaluated at $t =(2+h)\,s$ , measured in feet.

$h$ - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

$h = 0.1\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}$

$y(2.1) = 18.9\,ft$

$\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}$

$\bar v = -11\,\frac{ft}{s}$

$h = 0.01\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}$

$y(2.01) = 19.899\,ft$

$\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}$

$\bar v = -10.1\,\frac{ft}{s}$

$h = 0.001\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}$

$y(2.001) = 19.99\,ft$

$\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}$

$\bar v = -10\,\frac{ft}{s}$

b) The instantaneous velocity when $t = 2\,s$ can be obtained by using the following limit:

$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}$

$v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h$

$v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h$

$v(t) = 30-20\cdot t$

And we finally evaluate the instantaneous velocity at $t = 2\,s$ :

$v(2) = 30-20\cdot (2)$

$v(2) = -10\,\frac{ft}{s}$

The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

8 0

3 years ago

Write in radical form

Studentka2010 [4]

Step-by-step explanation:

I'll do one for you.

Using the formula for turning exponents into radicals

$(b) {}^{ \frac{x}{y} } = \sqrt[y]{b} {}^{x}$

where b is the base

This means that the numerator in the exponet form becomes the power under the radical in radical form and

the denominator in exponet form becomes the nth root in radical form.

For example 5,

$(5) {}^{ \frac{ - 3}{5} }$

That becomes in radical form

$\sqrt[5]{5 {}^{ - 3} }$

or if you want to write it using positive exponents

$\frac{1}{ \sqrt[5]{5 {}^{3} } }$

I'll do one more for you

For example 6,

$11 {}^{ \frac{4}{3} }$

That becomes in radical form

$\sqrt[3]{11 {}^{4} }$

4 0

3 years ago