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4 years ago

Math question down below

Mathematics

1 answer:

SCORPION-xisa [38]4 years ago

7 0

No is my answer, But i dont know if the is correct or not

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Bogdan [553]

B. your first number needs to be between 1-10

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3 years ago

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I need to know how to solve problem one. I tried it several different ways but I just can't seem to crack it. Can someone solve

grandymaker [24]

Either 100 ft or 50 ft

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4 years ago

The inches of rainfall in the rainforest is represented by the function f(x) = 3(x-2)+ 4. What is the average rate of change bet

nata0808 [166]

Answer:

The average rate of change of rainfall in the rainforest between 2nd year and 6th year = <u>3 inches</u>

Step-by-step explanation:

Given function representing inches of rainfall:

$f(x)=3(x-2)+4$

To find the average rate of change between the 2nd year and the 6th year.

Solution:

The average rate of change between interval $[a,b]$ is given as :

$\frac{f(b)-f(a)}{b-a}$

For the given function we need to find the average rate of change between 2nd year and 6th year. $[2,6]$

So, we have:

$f(2)=3(2-2)+4=3(0)+4=4$

$f(6)=3(6-2)+4=3(4)+4=12+4=16$

Thus, average rate of change will be:

$\frac{f(6)-f(2)}{6-2}$

⇒ $\frac{16-4}{6-2}$

⇒ $\frac{12}{4}$

⇒ $3$

Thus, the average rate of change of rainfall in the rainforest between 2nd year and 6th year = 3 inches

8 0

3 years ago

What is the slope of the line

STALIN [3.7K]

The answer you chose in the photo is correct !!

5 0

3 years ago

The accompanying data contains the depth (in kilometers) and magnitude, measured using the Richter Scale, of all earthquakes

Sunny_sXe [5.5K]

Answer:

Depth:

μ =20.2025 km

M = 15.625 km

Range = 47.15 km

σ ≈ 15.92 km

Q₁ = 5.7375 km

Q₃ = 34.6675 km

Magnitude:

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

Step-by-step explanation:

The given data are;

Depth ${}$ Magnitude

0.76 ${}$ 0.84

4.93 ${}$ 0.47

8.16 ${}$ 0.35

33.58 ${}$ 1.32

21.2 ${}$ 1.61

35.03 ${}$ 4.57

10.05 ${}$ 5.52

47.91 ${}$ 1.99

For the Depth, we have;

The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km

The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;

0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;

(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km

The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km

The Standard deviation of, σ, is given as follows;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}}$

Where;

$x_i$ = The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 )

N = The total number of data point = 8

Substituting, (using Microsoft Excel) we get;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-20.2025 \right )^{2} }{8}} \approx 15.92 \ km$

Q₁ = The first quartile = The (n + 1)/4th = term arranged in increasing order

Q₁ = The (8 + 1)/4th term = The 2.25th term = 4.93 + (8.16 - 4.93)×0.25) = 5.7375 km

Q₃ = The first quartile = The 3×(n + 1)/4th = term arranged in increasing order

Q₃ = The 3×(8 + 1)/4th term = The 6.75th term = 33.58 + 3×(35.03 - 33.58)×0.25) = 34.6675 km

For the magnitude, we have, using the same formulas and procedures as above;

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

4 0

4 years ago