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3 years ago

Solve for x. 4x−3+2x2−9=1x+3 −173 −5 133 193

Mathematics

1 answer:

AnnyKZ [126]3 years ago

3 0

Answer:

-1711005

Step-by-step explanation:

using BODMAS that implies you multiplication(M),before addition(A), then substraction(S) which comes last.

use the above rule to get the variable without letter only.

so ( 2*2)+(-3)=1

then 1-9=8, this part is for the left hand side which gives us 4x-8.

then the right hañd side 3-173-5133193 not including the variable with the letter x.now we have:1x-5133023.

so now

4x-8=1x-5133023

collect like terms

4x-1x=-5133023+8

3x=-5133015

divide both side by 3

x= -1711005

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What are the possible steps involved in solving the equation shown? Select three options. 3.5 + 1.2(6.3 – 7x) = 9.38 Add 3.5 and

motikmotik

Answer:

1) Distribute 1.2 to 6.3 and -7x

2)Combine 3.5 and 7.56

3)Subtract 11.06 from both sides

Step-by-step explanation:

3.5 + 1.2(6.3 - 7x) = 9.38

Distribute 1.2 to 6.3 and -7x

3.5 + 1.2* 6.3 - 1.2 * 7x = 9.38

3.5 + 7.56 - 8.4x = 9.38

Combine 3.5 and 7.56

11.06 - 8.4x = 9.38

Subtract 11.06 from both sides

11.06 - 8.4x -11.06 = 9.38 - 11.06

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To find solution:

Divide both sides by (-8.4)

-8.4x/-8.4 = -1.68/-8.4

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3 years ago

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3 years ago

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

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3 years ago

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Answer:

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2 years ago

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Answer:

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Step-by-step explanation:

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