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Nadusha1986 [10]
3 years ago
10

What is the area of parallelogram ABCD?

Mathematics
1 answer:
lukranit [14]3 years ago
3 0
The answer is b jdndnfjfnfnfnfnnd
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5x - 8y = 20<br> what are x and y? :)
Alex_Xolod [135]

Answer:

x could be 10

y could be 5

those are some possible answers.

8 0
2 years ago
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Or a very large set of data the measured mean is found to be 288.6 with a standard deviation of 21.2. assuming the data to be no
SashulF [63]

The 68-95-99.7 rule tells us 68% of the probability is between -1 standard deviation and +1 standard deviation from the mean. So we expect 75% corresponds to slightly more than 1 standard deviation.

Usually the unit normal tables don't report the area between -σ and σ but instead a cumulative probability, the area between -∞ and σ. 75% corresponds to 37.5% in each half so a cumulative probability of 50%+37.5%=87.5%. We look that up in the normal table and get σ=1.15.

So we expect 75% of normally distributed data to fall within μ-1.15σ and μ+1.15σ

That's 288.6 - 1.15(21.2) to 288.6 + 1.15(21.2)

Answer: 264.22 to 312.98


8 0
3 years ago
Need to know answer please
makvit [3.9K]

Answer:

IDK

Step-by-step explanation:

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3 0
3 years ago
The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th
vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by P(t)=1.394(1.006)^t

To find the derivative you need to:

\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t

To find the population growing at the start of 2014 we say t = 0

P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year

To find the population growing at the start of 2015 we say t = 1

P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year

To convert billion to million you multiple by 1000

P(0)' = 0.00833901 \:Billion/year \cdot 1000 = 8.34 \:Million/year \\P(1)' = 0.00838904 \:Billion/year \cdot 1000 = 8.39 \:Million/year

6 0
3 years ago
Y=2x2
yaroslaw [1]
C. Every y-value is twice the square of its x-value
3 0
3 years ago
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