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castortr0y [4]
3 years ago
11

-3=3x-3.solve for x?

Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0
Add 3 on both sides of the equation

3x=0

divide 3 on both sides of the equation

x=0
posledela3 years ago
6 0
Solution:
We have -3=3x-3
To solve for x, we would first isolate the variable x.
We start this by doing +3 to both sides, so that the equation still equals each other.
When you did that, you should have got
0=3x
Next, you divide each side by 3 to isolate x completely to get
 x=0
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2 years ago
The sum of an arithmeit cprogression consisting of 20 positive integer terms with positive common difference is equal to 2020. (
Rufina [12.5K]

Answer:

a) possible progressions are 5

b) the smallest and largest possible values of the first term are 16 and 82

Step-by-step explanation:

<u>Sum of terms:</u>

  • Sₙ = n/2(a₁ + aₙ) = n/2(2a₁ + (n-1)d)
  • S₂₀ = 20/2(2a₁ + 19d) = 10(2a₁ + 19d)
  • 2020 = 10(2a₁ + 19d)
  • 202 = 2a₁ + 19d

<u>In order a₁ to be an integer, d must be even number, so d = 2k</u>

  • 202 = 2a₁ + 38k
  • 101 = a₁ + 19k

<u>Possible values of k= 1,2,3,4,5</u>

  • k = 1 ⇒ a₁ = 101 - 19 = 82
  • k = 2 ⇒ a₁ = 101 - 38 = 63
  • k = 3 ⇒ a₁ = 101 - 57 = 44
  • k = 4 ⇒ a₁ = 101 - 76 = 25
  • k = 5 ⇒ a₁ = 101 - 95 = 16

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3 years ago
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3 years ago
The travel-to-work time for residents in Ventura County is unknown. Assume the population variance is 39. How large should a sam
katen-ka-za [31]

Answer:

A sample size of 128 is needed.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.93}{2} = 0.035

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.035 = 0.965, so z = 1.81

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population(square root of the variance) and n is the size of the sample.

How large should a sample be if the margin of error is 1 minute for a 93% confidence interval

We need a sample size of n, which is found when M = 1. We have that \sigma = \sqrt{39}. So

M = z*\frac{\sigma}{\sqrt{n}}

1 = 1.81*\frac{\sqrt{39}}{\sqrt{n}}

\sqrt{n} = 1.81\sqrt{39}

(\sqrt{n})^2 = (1.81\sqrt{39})^2

n = 127.8

Rounding up

A sample size of 128 is needed.

7 0
3 years ago
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