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slamgirl [31]
3 years ago
10

(10+12x)(15x+1) please help me

Mathematics
1 answer:
MrRa [10]3 years ago
4 0
180x^2+162x+10


Please mark me brainless
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Can you solve this? Please ( 15+ points)
Vera_Pavlovna [14]

Answer:

112+24x

Step-by-step explanation:

2(16+(10*4)+(x*12))

solve the inner bracket

2(16+40+12x)

solve this bracket

2(56+12x)

solve this now

112+24x

6 0
2 years ago
Read 2 more answers
Find the length of side a<br>will mark brainly
Mekhanik [1.2K]

Answer:

5

Step-by-step explanation:use the cube to count(cube is in the corner.)

5 0
3 years ago
Represent Jacey wrote an equation to
Anika [276]

Answer:

b is the angle being measured while a is the angle at a point with b

Step-by-step explanation:

The only time we use 360 for calculating an angle is when the two angles are at a point as both of them assuming a and b sum up to 360°.

Therefore a + b = 360

and if b is being measured, b = 360 -a

3 0
3 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
Please help! thanks​
anygoal [31]

1 =1,1,4. 5=13,16

2 =1,4. 6=23,3

3 =18,55. 7= 25,8

4 =3,16.

ps , meanes fraction

7 0
2 years ago
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