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vekshin1
3 years ago
13

What is the estimate of 204878

Mathematics
1 answer:
Yuki888 [10]3 years ago
4 0
The the nearest hundred thousand 200000
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Three times the sum of a number and 21 is at most -26
Anna007 [38]
It would be: 3(x + 21) ≤ -26
3x + 21 ≤ -26
3x ≤ -47
x < -47/3

In short, Your Answer would be -47/3 or 15.67

Hope this helps!

8 0
3 years ago
what is an equation for the line that passes through (-1,-4) and (1,4)? Write the equation in slope-intercept form.
hichkok12 [17]
To find the slope use the formula Y2-Y1 --------- X2-X1 (-1,-4) (1,4) 4- -4 = 8 ------------- 1- -1 = 2 8/2=4 This will be your slope Y=4x Now we must figure out the y-intercept using the formula Y-y1=slope(x-x1) Y- -4= 4(x- -1) Y+4= 4x + 4 The 4 distributes to the x and -1 Y=4x +4 -4 Y= 4x This is your answer in slope intercept form. P.S if you have a number minus a negative number it makes it a positive number
7 0
4 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 <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b) <em>n</em> = 28.09

Step-by-step explanation:

The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.

It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of <em>Y</em>ₙ are:

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

(a)

For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

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

Compute the probability that <em>Y</em>₂₀ 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 <em>z </em>table for probability.

Thus, the probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b)

It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

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

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

Compute the value of <em>n</em> 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 <em>n</em> = 28.09.

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Mrs.Solar modelded Contractor 1's fee formula with the equation y = 50x 200.
KonstantinChe [14]
I WOULD DO THE EASIEST, A.

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