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Reptile [31]
3 years ago
15

Find the value of x such that the data set 31.7,42.8,26.4, x has a mean of 35

Mathematics
2 answers:
Komok [63]3 years ago
7 0
The answer is 39.1
(31.7+42.8+26.4+x)/4=35
Multiply both sides by 4.
100.9 +x=140
Subtract 100.9 from both sides you get 39.1
expeople1 [14]3 years ago
5 0

Answer:  The required value of x is 39.1.

Step-by-step explanation:  Given that the mean of the following data set is 35.

31.7,~42.8,~26.4,~x.

We are to find the value of x.

To find the mean of a data set, we need to add all the values together and divide by the number of values in the set.

The sum of all the values in the given data is given by

31.7+42.8+26.4+x\\\\=100.9+x.

And the total number of values in the given data = 4.

Therefore, the MEAN of the data is

\dfrac{100.9+x}{4}=35\\\\\\\Rightarrow 100.9+x=140\\\\\Rightarrow x=140-100.9\\\\\Rightarrow x=39.1.

Thus, the required value of x is 39.1.

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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
Viktor [21]

Answer:

So Philip made 5 bracelets and 4 necklaces.

Step-by-step explanation:

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

Since we have a total of 9 bracelets and necklaces,

x + y = 9 (1)

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.

So, the total length for both is 8x + 20y. Since the total length of cord used is 120 inches,

8x + 20y = 120 (2)

Simplifying it we have

2x + 5y = 30  (3).

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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Georgia [21]

Answer:

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Step-by-step explanation:

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Vinil7 [7]

Solution-

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8 0
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