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sweet-ann [11.9K]
3 years ago
6

Triangle IJK, with vertices I(3,-8), J(9,-7), and K(5,-4), is drawn inside a rectangle, as shown below. What is the area, in squ

are units, of triangle IJK?

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
4 0

Answer:

11 sq units

Step-by-step explanation:

(6×4) - ½(6×1 + 4×3 + 2×4)

= 24 - ½(6+12+8)

= 24 - ½(26)

= 24-13

= 11 sq units

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Enter the y coordinate of the solution to this system of equations. <br> -2x+3y=-6<br> 5x-6y=15
julia-pushkina [17]
Y coordinate on solving both equations comes out to be 0
-2x+3y=-6
3y = -6+2x
put the value of 3y in equation 2nd
5x-2(-6+2x) =15
5x+12-4x = 15
x=3
put value of X in 3y = -6+2x
3y = -6+2*3 = -6+6 = 0
thus y = 0
7 0
3 years ago
1) 28 + 3[{12 + 2 (14~4) = 5} - 6] 3​
nasty-shy [4]

Answer:

Step-by-step explanation:

whats ur question?

8 0
2 years ago
According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann
mihalych1998 [28]

Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of consecutive time intervals in which the water supply remains below a critical value <em>y₀</em>.

The random variable <em>Y</em> follows a Geometric distribution with parameter <em>p</em> = 0.409<em>.</em>

The probability mass function of a Geometric distribution is:

P(Y=y)=(1-p)^{y}p;\ y=0,12...

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) =  P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

              =(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable <em>Y</em> as follows:

\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445

Compute the standard deviation of the random variable <em>Y</em> as follows:

\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

                    = P (Y ≥ 3.325)

                    = P (Y ≥ 3)

                    = 1 - P (Y < 3)

                    = 1 - P (X = 0) - P (X = 1) - P (X = 2)

                    =1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

6 0
3 years ago
An architect planned to construct two similar stone pyramid structures in a park. The
kondaur [170]

Answer:

1) Change the length of side AB to 2 feet

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To find out what should be the correct length of AB that she should change to, set up the proportion showing the ratio of 2 corresponding lengths of both structures. Thus:

\frac{PR}{AC} = \frac{PQ}{AB}

We will assume AB is unknown.

PR = 7.5 ft

AC = 2.5 ft

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Plug in the values into the equation

\frac{7.5}{2.5} = \frac{6}{AB}

Cross multiply

AB*7.5 = 6*2.5

AB*7.5 = 15

Divide both sides by 7.5

AB = 2

The architect should change the length of AB to 2 ft

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the answer id d because it's 3 z's so you times that by 4 and you get d.4

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3 years ago
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