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Wewaii [24]
3 years ago
8

Rewrite this equation in function notation. U is the dependent variable. -8v+71=u

Mathematics
2 answers:
Lady bird [3.3K]3 years ago
7 0
My sister is learning this too ,ill help you in a few minutes
dimulka [17.4K]3 years ago
3 0
So basically rearrange the equation so that v is the subject.
Thus,
-8v = u - 71
v = [(u - 7)/-8]
hope this helped :)
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AnnZ [28]

Answer:

The second option: "Even though Antartica is cold and has snow, it is a desert."

Step-by-step explanation:

This is the answer because by reading the paragraph, you will see that the main idea is stated in the first few sentences.

5 0
3 years ago
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At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t
grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c) The expected value of the temperature increase is 17.5°C.

Step-by-step explanation:

Let <em>X</em> = temperature increase.

The random variable <em>X</em> follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of <em>X</em> is:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

P(X

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

P(20

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable <em>X</em> as follows:

E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

Thus, the expected value of the temperature increase is 17.5°C.

7 0
3 years ago
Is algebra.
Nostrana [21]

Answer:

1) b   2) b

Step-by-step explanation:

1) Both expressions have (x+6). Rearrange them and you'll have one expression as (x+6) and the other as (5ab-4).

2) (4b - 7x)(a + b) factors to be 4ba + 8b - 7ax - 14x, which can be rearranged to 4ab - 7ax + 8b - 14x

3 0
2 years ago
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A blue whale traveled 31 1/2 miles in 2 1/4 hours. What was the whales rate in yards per hour ( 1 mile = 1,760 yards
tekilochka [14]

Answer: 25594.018691648 yards per hour

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Step-by-step explanation:

6 0
3 years ago
A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between
sattari [20]

Answer:

a)

B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)

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Step-by-step explanation:

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B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

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b) rate of increase after one day.

We put t = 1

B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09

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8 0
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