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2 years ago

Solve the following: a) 4x 3= 2x + 7 b) 2x+6= 7x14

Mathematics

1 answer:

nikdorinn [45]2 years ago

7 0

<h2>Correct question : 4x - 3 = 2x + 7 </h2>

<h2>Solution : 4x - 3 = 2x + 7 </h2>

$4x - 2x = 7 + 3 \\ \\ 2x = 10 \\ \\ x = \frac{10}{2} \\ \\ x = 5.$

<h2>Correct Question : 2x + 6 = 7x - 14</h2>

<h2>Solution : 2x + 6 = 7x - 14</h2>

$2x - 7x = - 14 - 6 \\ \\ - 5x = - 20 \\ \\ x = \frac{20}{5} \\ \\ x = 4.$

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Simplify the expression. -5a 3b 8a

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Here is u answer -5a4b8

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3 years ago

Find The sum of the first 10 terms 8,20,32,44

nikklg [1K]

Answer:

S₁₀=620

Step-by-step explanation:

This is an arithmetic sequence with a common difference of d=12

The formula for an arithmetic series is Sₙ=(n/2)(a₁+aₙ) where aₙ is the nth term and a₁ is the first term.

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3 years ago

No link need right answer 100 points

maks197457 [2]

Answer:

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Step-by-step explanation:

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4 years ago

Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below. (Round all an

Gelneren [198K]

Answer:

(a) P (Z < 2.36) = 0.9909 (b) P (Z > 2.36) = 0.0091

(e) P (-0.77< Z > -0.55) = 0.0705 (f) P (Z > 3) = 0.0014

(g) P (Z > -3.28) = 0.9995 (h) P (Z < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, $X \sim N (\mu, \sigma^{2})$ , then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, $Z \sim N (0, 1)$ .

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean. The z-scores are standardized scores.

The distribution of these z-scores is known as the standard normal distribution.

(a)

Compute the value of P (Z < 2.36) as follows:

P (Z < 2.36) = 0.99086

≈ 0.9909

Thus, the value of P (Z < 2.36) is 0.9909.

(b)

Compute the value of P (Z > 2.36) as follows:

P (Z > 2.36) = 1 - P (Z < 2.36)

= 1 - 0.99086

= 0.00914

≈ 0.0091

Thus, the value of P (Z > 2.36) is 0.0091.

(c)

Compute the value of P (Z < -1.22) as follows:

P (Z < -1.22) = 0.11123

≈ 0.1112

Thus, the value of P (Z < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < Z > 3.35) as follows:

P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)

= 0.99960 - 0.87076

= 0.12884

≈ 0.1288

Thus, the value of P (1.13 < Z > 3.35) is 0.1288.

(e)

Compute the value of P (-0.77< Z > -0.55) as follows:

P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)

= 0.29116 - 0.22065

= 0.07051

≈ 0.0705

Thus, the value of P (-0.77< Z > -0.55) is 0.0705.

(f)

Compute the value of P (Z > 3) as follows:

P (Z > 3) = 1 - P (Z < 3)

= 1 - 0.99865

= 0.00135

≈ 0.0014

Thus, the value of P (Z > 3) is 0.0014.

(g)

Compute the value of P (Z > -3.28) as follows:

P (Z > -3.28) = P (Z < 3.28)

= 0.99948

≈ 0.9995

Thus, the value of P (Z > -3.28) is 0.9995.

(h)

Compute the value of P (Z < 4.98) as follows:

P (Z < 4.98) = 0.99999

≈ 0.9999

Thus, the value of P (Z < 4.98) is 0.9999.

**Use the z-table for the probabilities.

3 0

3 years ago