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Whitepunk [10]
3 years ago
8

Ms more wrote the following on the blackboard and asked her students to write it in fraction form:

Mathematics
1 answer:
TiliK225 [7]3 years ago
5 0
660.011/1 would be equivalent to that expression
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Hailey wants to put a skirt around her desk to hide all her computer equipment and cords. She measured her desk, found the dimen
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Answer:

C.\ Perimeter = 48 * 18

Step-by-step explanation:

Given

See attachment for her sketch

Required

Which equation do not represent the perimeter

From the attached sketch:

L =18in --- Length

W =48in --- Width

Perimeter (P) is calculated as:

P = 2 * (L + W)

This gives:

P = 2 * (18 + 48) ---- This represents (A)

Open bracket

P = 2 * 18 + 2*48 ---- This represents (B)

In algebra:

2 * a means a+ a

So, the expression becomes

P = 18 + 18 +48 + 48 --- This represents (D)

<em>This implies that (C) does not represent the perimeter</em>

4 0
2 years ago
NEED HELP 10 POINTS
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48.8 megabytes. When the download is 17.5% complete, have been downloaded 48.8 megabytes.

The key to solve this problem is using the rule of three.

If 279Mb -------------> 100%

      x      <-------------   17.5%

x = [(279 megabytes)(17.5)]/100

x = 48.825 megabytes

Round to the nearest tenth 48.8 megabytes.

6 0
3 years ago
What is 2 2 3 ÷ 2 3 ?
Ivahew [28]

Answer:

9.695652174 or 9.69 or 9.70

Step-by-step explanation:

223÷23

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To fit in an existing frame, the length, x, of a piece of glass must be longer than 12 cm but not longer than 12.2 cm. Which ine
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The answer is A) <span>12 < x ≤ 12.2</span>
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The obtained answers for the given frequency distribution are:

(a) The formula for the mean in sigma notation is \bar x =\frac{1}{n}  \sum X_i where n is the number of observations; X_i are the n observations.

The mean for the given monthly plan price is $16.1.

(b) The frequency distribution for given data is {$9.99 - 2; $10 - 5; $12 - 1; $12.75 - 2; $14.99 - 6; $20 - 4; $25 - 5}

(c) The formula for the mean using the frequency distribution table is \bar x = \frac{1}{N}\sum f_ix_i where N =\sum f_i and on applying this formula for the given data, the mean is $16.1.

(d) The median for the given data is m_e = 14.99, and the mode for the given data is $14.99

<h3>What are the mean, median, and mode for a frequency distribution?</h3>

The frequency distribution has sample observations x_i and frequencies f_i.

Then, the mean is calculated by

\bar x = \frac{1}{N}\sum f_ix_i

Where N =\sum f_i (Sum of frequencies)

The median is calculated by

m_e=\left \{ {{x_{k}} \ if \ n = 2k+1 \atop {\frac{x_{k}+x_{k+1}}{2}} \ if \ n =2k} \right.

The mode is calculated by

Mode = highest frequency value

<h3>Calculation:</h3>

The given list of data is

{$14.99, $12.75, $14.99, $14.99, $9.99, $25, $25, $10, $14.99, $10, $20, $10, $20, $14.99, $10, $25, $20, $12, $14.99, $25, $25, $20, $12.75, $10, $9.99}

(a) Formula for the mean using sigma notation and use it to calculate the mean:

The formula for the mean is

\bar x =\frac{1}{n}  \sum X_i

Where n = 25; X_i - n observations

On substituting,

Mean \bar x

=1/25(14.99+12.75+14.99+14.99+9.99+25+25+10+14.99+10+20+10+20+14.99+10+25+20+12+14.99+25+25+20+12.75+10+9.99)

= 1/25(402.42)

= 16.09 ≅ 16.1

(b) Constructing a frequency distribution for the data:

Cost - frequency - cumulative frequency

$9.99 - 2 - 2

$10 - 5 - 7

$12 - 1 - 8

$12.75 - 2 - 10

$14.99 - 6 - 16

$20 - 4 - 20

$25 - 5 - 25

Sum of frequencies N = 25;

(c) Using frequency distribution, calculating the mean:

The formula for finding the mean using frequency distribution is

\bar x = \frac{1}{N}\sum f_ix_i

Where N = 25;

On substituting,

\bar x<em> </em>= 1/25 (2 × 9.99 + 5 × 10 + 1 × 12 + 2 × 12.75 + 6 × 14.99 + 4 × 20 + 5 × 25)

  = 1/25 (402.42)

  = 16.09 ≅ 16.1

Therefore, the mean is the same as the mean obtained in option (a).

(d) Calculating the median and the mode:

Since N = 25(odd) i.e., 2· 12 + 1; k = (12 + 1)th term = 13th term

So,  the median m_e = 14.99. (frequency at 13th term)

Since the highest frequency is 6 occurred by the cost is $14.99,

Mode = 14.99

Learn more about frequency distribution here:

brainly.com/question/27820465

#SPJ9

7 0
2 years ago
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