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Sunny_sXe [5.5K]
3 years ago
13

Please help I will mark as brianliest

Mathematics
1 answer:
shutvik [7]3 years ago
8 0

Answer:

x = -4

Step-by-step explanation:

-5x-4x-2 = -8x+2 is the only one with a solution.

-5x-4x-2 = -8x+2

-x-2 = 2

-x = 4

x = -4

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The figure shows three exterior angles of triangle ABC. Show that the measure of angle 1 plus the measure of angle 2 plus the me
ivolga24 [154]

Step-by-step explanation:

Let the angle near point A be A.

Let the angle near point B be B.

Let the angle near point C be C.

Let the external angle near point A be extA.

Let the external angle near point B be extB.

Let the external angle near point C be extC.

According to exterior angle property,exterior angle of an angle is the sum of other two angles.

So,

extA=B+C\\extB=A+C\\extC=A+B

According to angle sum property,the sum of angles in the triangle add to 180

extA+extB+extC=B+C+A+C+A+B=2(A+B+C)=2\times 180=360^{0}

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3 years ago
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What's 5 + 5 × 10 - 88 + 200 - 1​
Mars2501 [29]

Answer:

=50x+116

Step-by-step explanation:

5+5x(10)−88+200−1

=5+50x+−88+200+−1

Combine Like Terms:

=5+50x+−88+200+−1

=(50x)+(5+−88+200+−1)

=50x+116

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3 years ago
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Find the nth term of the following sequence: 7, 13, 19, 25,
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The answer of the next term is 31
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The ratio of boys to girls at a daycare is 5 to 4 which of these cannot be the total number of kids at the daycare ?
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The score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.
Artemon [7]

Answer:

P(X \geq 74) = 0.3707

Step-by-step explanation:

We are given that the score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Let X = Score of golfers

So, X ~ N(\mu=73,\sigma^{2}=3^{2})

The z score probability distribution is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = population mean = 73

           \sigma = standard deviation = 3

So, the probability that the score of golfer is at least 74 is given by = P(X \geq 74)

 P(X \geq 74) = P( \frac{X-\mu}{\sigma} \geq \frac{74-73}{3} ) = P(Z \geq 0.33) = 1 - P(Z < 0.33)

                                               =  1 - 0.62930 = 0.3707                  

Therefore, the probability that the score of golfer is at least 74 is 0.3707 .

3 0
3 years ago
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