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3 years ago

Find the slope of the line that passes through (4,-5) and (-2,9)

Mathematics

2 answers:

Delvig [45]3 years ago

6 0

The answer is C
Hope this helps

iogann1982 [59]3 years ago

3 0

Answer:

sorry i dont know

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Please help I got stuck on this on thank you

svetlana [45]

Answer is B. 3

2/3 x 3 = 6/3 = 2 cups

7 0

3 years ago

(-5,-2) (1,2) find the point-slope equation produced by the points

ANEK [815]

Answer:

M=2+2/1+5=2/3 is the slope

Step-by-step explanation:

8 0

3 years ago

A rectangular pool contains 920 cubic meters of water. the pool is .4 meter deep and 100 meters wide. How long is the pool?

LekaFEV [45]

Hi there,

Sol:
Length of the swimming pool = 260 m
Breadth of the swimming pool = 140 m

Let the height of the water in the swimming pool after pouring water be h meters.

Volume of the water poured = 54600 m3

l x b x h = 54600
260 x 140 x h = 54600
h = (54600) / (260 x 140)
h = 1.5 m = 150 cm

Hope it helps

5 0

3 years ago

Find thetwo consecutive integers whose product is 24

malfutka [58]

Answer:

There is no solution to this equation that is an integer for x.

Step-by-step explanation:

x is the first integer, and x+1 the second

$x(x+1) = 24\\x^{2} + x - 24 =0\\$

$x = \frac{-1+\sqrt{1^{2}+4*24}}{2} =\frac{-1+\sqrt{97}}{2}\\x = \frac{-1-\sqrt{1^{2}+4*24}}{2} =\frac{-1-\sqrt{97}}{2}$

5 0

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.

8 0

3 years ago