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3 years ago

-9=3x+6 what is the value of x?

Mathematics

2 answers:

Mkey [24]3 years ago

8 0

Answer:

-5

Step-by-step explanation:

3x + 6 = -9

3x = -15

x = -5

cluponka [151]3 years ago

5 0

Answer:

$\boxed{ \bold { \huge{ \boxed{ \sf{x = - 5}}}}}$

Step-by-step explanation:

$\sf{ - 9 = 3x + 6}$

Swap the sides of the equation

$\longrightarrow{ \sf{3x + 6 = - 9}}$

Move 6 to right hand side and change its sign

$\longrightarrow{ \sf{3x = - 9 - 6}}$

The negative integers are always added but possess the negative ( - ) sign

$\longrightarrow{ \sf{3x = - 15}}$

Divide both sides by 3

$\longrightarrow{ \sf{ \frac{3x}{3} = \frac{ - 15}{3} }}$

Calculate

$\longrightarrow{ \sf{x = - 5}}$

Hope I helped!

Best regards! :D

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Which pair of numbers, if included in this set, would not change the median?

sattari [20]

Answer:

Step-by-step explanation:

Order the set first

27, 28, 35, 37, 43, 47

Median now is: 36 (which is the mean between 35 and 37, the two central numbers )

If you add A) 35,50 then you will still have to compute the mean between 35 and 37 in order to find the median, because you have:

27,28,35,35,37,43,47,50

7 0

4 years ago

Read 2 more answers

A regional automobile dealership sent out fliers to prospective customers indicating that they had already won one of three diff

igor_vitrenko [27]

Answer:

(a) 31101

(b) $5.65

Step-by-step explanation:

The complete question is:

A regional automobile dealership sent out fliers to prospective customers indicating that they had already won one of three different prizes: an automobile valued at $20,000, a $125 gas card, or a $5 shopping card. To claim his or her prize, a prospective print on the back of the flier listed the probabilities of winning. The chance of winning the car was 1 out of 31,101, the chance of winning the gas card was 1 out of 31,101, and the chance of winning the shopping card was 31, 099 out of 31,101.

Solution:

The information provided is as follows:

$P(\text{Car})=\frac{1}{31101}\\\\P(\text{Gas Card})=\frac{1}{31101}\\\\P(\text{Shopping Card})=\frac{31099}{31101}$

(a)

The number of fliers the automobile dealership sent out is, <em>n</em> = 31,101.

This is because the probability of winning any of the three prize is out of 31,101.

(b)

Compute the expected value of the prize won by a prospective customer receiving a flier as follows:

$E(X)=\sum x\times P(X=x)$

$=[20000\times \frac{1}{31101}]+[125\times \frac{1}{31101}]+[5\times \frac{31099}{31101}]\\\\=0.6431+0.0040+4.9997\\\\=5.6468\\\\\approx 5.65$

Thus, the expected value of the prize won by a prospective customer receiving a flier is $5.65.

(c)

Compute the standard deviation of the prize won by a prospective customer receiving a flier as follows:

$SD(X)=\sqrt{V(X)}=\sqrt{E(X^{2}-(E(X))^{2}}$

$=\sqrt{[(20000)^{2}\times \frac{1}{31101}+(125)^{2}\times \frac{1}{31101}+(5)^{2}\times \frac{31099}{31101}]-(5.65)^{2}}\\\\=\sqrt{12854.9011}\\\\=113.37947\\\\\approx 113.38$