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saw5 [17]
3 years ago
7

Roberta has collected a set of data and calculated the mean to be 34. The set contains 75 numbers, but Roberta gets an email tel

ling her to add the following 5 numbers to her data: {23, 26, 32, 33, 40}
To the nearest tenth, what is the new mean for her data?


A) 27.8


B) 30.8


C) 33.8


D) 36.8
Mathematics
2 answers:
Alexus [3.1K]3 years ago
7 0

Answer:

C) New Mean = 33.8

Step-by-step explanation:

Mean of collected set of data = 34

Total number of observations = 75

Mean = \frac{\text{Sum of all observation}}{\text{Number of observations}}

Thus, we can  calculate the sum of observations as : Mean × Total number of observations = 2550

5 new numbers were added to data: 23, 26, 32, 33, 40

New sum of observations = 2550 + 23 + 26 + 32 + 33 + 40 = 2704

New total number of observations = 75 + 5 = 80

New Mean = \frac{\text{New sum of all observation}}{\text{New number of observations}} = \frac{2704}{80} = 33.8

Thus, option C) 33.8 is the new mean.

IRINA_888 [86]3 years ago
6 0
30.8 add it all up and divided how ever many number there are
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Philip made a total of 9 bracelets and necklaces from 120 inches of cord. He used 8 inches of cord for each bracelet and 20 inch
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Answer:

So Philip made 5 bracelets and 4 necklaces.

Step-by-step explanation:

Let x = number of bracelets and y = number of necklaces.

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Also, we have 8 inches of cord for each bracelet and 20 inches of cord for each necklace, then the total length for the bracelet is 8x and that for the necklace is 20y.

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Simplifying it we have

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Writing equations (1) and (3) in matrix form, we have

\left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}9\\30\end{array}\right]

Using Cramer's rule to solve for x and y,

x = det \left[\begin{array}{ccc}9&1\\30&5\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\

x = (9 × 5 - 30 × 1) ÷ (1 × 5 - 1 × 2)

x = (45 - 30) ÷ (5 - 2)

x = 15 ÷ 3

x = 5

y = det \left[\begin{array}{ccc}1&9\\2&30\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\

y = (30 × 1 - 9 × 2) ÷ (1 × 5 - 1 × 2)

y = (30 - 18) ÷ (5 - 2)

y = 12 ÷ 3

y = 4

So Philip made 5 bracelets and 4 necklaces.

3 0
3 years ago
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