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fredd [130]
2 years ago
6

Please please help ASAP I will mark brainly

Mathematics
1 answer:
Semmy [17]2 years ago
6 0
0647 is the answer hope I helped you out in any way
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Complete the following direct proof:
anzhelika [568]
B. I hope this help please let me know if it's correct.
4 0
2 years ago
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7. 60% of 90<br> Find the percent of the quantity
NeTakaya

Answer:

54

Step-by-step explanation:

60% of 90 = 60/100 × 90

= 0.6 × 90

= 54

Therefore, 60% of 90 is 54.

4 0
3 years ago
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
Plz help I need the answer <br><br> You have to solve for the variable
serious [3.7K]
X=10 let me know if you need the work
6 0
3 years ago
Read 2 more answers
The diagram shows a logo​
Charra [1.4K]

Answer/Step-by-step explanation:

✔️Find EC using Cosine Rule:

EC² = DC² + DE² - 2*DC*DE*cos(D)

EC² = 27² + 14² - 2*27*14*cos(32)

EC² = 925 - 756*cos(32)

EC² = 283.875639

EC = √283.875639

EC = 16.85 cm

✔️Find the area of ∆DCE:

Area = ½*14*27*sin(32)

Area of ∆DCE = 100.15 cm²

✔️Since ∆DCE and ∆ABE are congruent, therefore,

Area of ∆ABE = 100.15 cm²

✔️Find the area of the sector:

Area of sector = 105/360*π*16.85²

Area = 260.16 cm² (nearest tenth)

✔️Therefore,

Area of the logo = 100.15 + 100.15 + 260.16 = 460.46 ≈ 460 cm² (to 2 S.F)

5 0
3 years ago
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