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natima [27]
2 years ago
8

Solve for x: -3(x+5) = -9 Ox= -5 Ox=-2 Ox=2 Ox= 1/3

Mathematics
1 answer:
netineya [11]2 years ago
8 0
Answer X=-2

-3x - 15=9
-3x= -9 + 15
-9 + 15= 6
-3x = 6
x= -6/3
x= -2
answer x=-2
sorry if this wasn’t the best explanation i’m kinda rushing rn bc of class
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For which sample size (n) and sample proportion (p) can a normal curve be
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Answer:

Option c.

Step-by-step explanation:

Using the normal curve to approximate a sampling distribution:

For a sample size n and a proportion n, the normal curve can be used if:

np \geq 10 and n(1-p) \geq 10

Option a:

np = 35*0.8 = 28 > 10

n(1-p) = 35*0.2 = 7 < 10

So option a cannot be used.

Option b:

np = 65*0.9 = 58.5 > 10

n(1-p) = 65*0.1 = 6.5 < 10

So option b cannot be used.

Option c:

np = 65*0.8 = 52 > 10

n(1-p) = 65*0.2 = 13 > 10

So option c can be used, and is the answer

Option d:

np = 35*0.9 = 31.5 > 10

n(1-p) = 35*0.1 = 3.5 < 10

So option d cannot be used.

The answer is given by option c.

7 0
3 years ago
Given congruent triangles name the corresponding sides and corresponding angles
Alborosie

Answer:

See explanation

Step-by-step explanation:

Given \triangle ABC\cong \triangle ADC

According to the order of the vertices,

  • side AB in triangle ABC (the first and the second vertices) is congruent to side AD in triangle ADC (the first and the second vertices);
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  • side AC in triangle ABC (the first and the third vertices) is congruent to side AC in triangle ADC (the first and the third vertices);
  • angle BAC in triangle ABC is congruent to angle DAC in triangle ADC (the first vertex in each triangle is in the middle when naming the angles);
  • angle ABC in triangle ABC is congruent to angle ADC in triangle ADC (the second vertex in each triangle is in the middle when naming the angles);
  • angle BCA in triangle ABC is congruent to angle DCA in triangle ADC (the third vertex in each triangle is in the middle when naming the angles);
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Your welcome

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Hello, Welcome to Brainly I am papaguy expert User I will you if you have any questions so far you have on.


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