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musickatia [10]
3 years ago
12

HI. Can you help???

Mathematics
1 answer:
Maslowich3 years ago
3 0

Answer:

The maximum rate of change is between second 15 and second 19.

Step-by-step explanation:

See the attached table.

The table shows the total distance a spider traveled after a light was turned on.

Now, the average rate of change between two consecutive times in the table is given by  

= \frac{\textrm {Total change in Distance in feet}}{\textrm {Total change in Time in seconds }}

Therefore, between second 5 and second 11, the average rate of change = \frac{7 - 4}{11 - 5} = 0.5

Between second 11 and second 15, the average rate of change = \frac{13 - 7}{15 - 11} = 1.5

Between second 15 and second 19, the average rate of change = \frac{20 - 13}{19 - 15} = 1.75

Between second 19 and second 28, the average rate of change = \frac{24 - 20}{28 - 19} = 0.44

Therefore, the maximum rate of change is between second 15 and second 19. (Answer)

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Studentka2010 [4]

Answer:

11

Step-by-step explanation:

3 0
3 years ago
In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi
Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n  = 50

sample mean \overline x = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

X \sim N ( \mu \sigma)

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})

P(X>140) = 1 - P( Z\leq0.42)

From z tables,

P(X>140) = 1 - 0.6628

P(X>140) = 0.3372

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

Y \sim Binomial (np)

Y \sim Binomial (50,0.3372)

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49}  + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +...  + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}

P(Y ≥ 15) = 0.763

7 0
3 years ago
Simplify your answer to the previous part and enter a differential equation in terms of the dependent variable xx satisfied by x
rusak2 [61]

Answer:

x'-5x=0, or x''-25x=0, or x'''-125x=0

Step-by-step explanation:

The function x(t)=e^{5t} is infinitely differentiable, so it satisfies a infinite number of differential equations. The required answer depends on your previous part, so I will describe a general procedure to obtain the equations.

Using rules of differentiation, we obtain that x'(t)=5e^{5t}=5x \text{ then }x'-5x=0. Differentiate again to obtain, x''(t)=25e^{5t}=25x=5x' \text{ then }x''-25x=0=x''-5x'. Repeating this process, x'''(t)=125e^{5t}=125x=25x' \text{ then }x'''-125x=0=x'''-25x'.

This can repeated infinitely, so it is possible to obtain a differential equation of order n. The key is to differentiate the required number of times and write the equation in terms of x.

7 0
3 years ago
What's the answer to 3s-4=1-3s
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6 0
3 years ago
Read 2 more answers
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3 years ago
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