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Helga [31]
3 years ago
9

I need help with this and fast if possible

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0

Answer:

<em>The answer is Hence Proved</em>

Step-by-step explanation:

       Given that CB║ED ,  CB ≅ ED

       To prove Δ CBF ≅ Δ EDF

  •       CB ≅ ED   ( Given )

        This means that the length of CB is equal to ED

        As CB║ED The following conditions satisfies when a transversal cut

        two parallel lines

  •    ∠ EDF = ∠ FBC   ( Alternate interior points )
  •    ∠ DEF = ∠ FCB   ( Alternate interior points )

       ∴ Δ CBF ≅ Δ EDF    ( By ASA criterion)

        The Δ CBF is congruent to Δ EDF By ASA criterion .

    <em>    Hence proved </em>

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) Ms. Pimentel went on a trip with the Glee Club. The trip was $365. Included in that price are $105 for a plane ticket, the cos
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A. $25
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Jack is holding nickels and dimes. He has 4 more dimes than nickels. He has a total of $.70 in his hand. How many of each coin d
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A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a
natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, .   .   .  , x₂₃ be the actual data observed by the student

The sample means  = x₁  +  x₂  +  .   .   .  , x₂₃ / 23

= \frac{x_1 +x_2 +...x_2_3}{23}

= 2.3hr

⇒\sum xi =2.3 \times 23 = 52.9hrs

let x₁ , x₂, .   .   .  , x₂₃  arranged in ascending order

Then x₂₃ was 10  and has been changed to 14

i.e x₂₃ increase to 4

Sample mean  = \frac{x_1 +x_2 +...x_2_3}{23}

\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47

therefore, the new sample mean is 2.47

2) For the old data set

the median is x_1_2(th) values

[\frac{n +1}{2} ]^t^h value

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = [{x_1_2]^t^h value remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)

The new sample standard sample deviation is calculated as

= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)

Now, when we compare (1) and (2)  the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)²  ∑ (xi - 2.47)²

Therefore , the standard deviation increases

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2x+3=2

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