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2 years ago

Izumi is running on a quarter-mile oval track. After running 110 yards, his coach records his time as 16 seconds.

Mathematics

1 answer:

pickupchik [31]2 years ago

3 0

Answer:

14.04 miles per hour

Step-by-step explanation:

The problem is asking for Izumi's speed. The formula of speed is:

s = $\frac{d}{t}$ , where "s" means speed, "d" means distance and "t" means time.

The problem is also asking for the unit<u> miles per hou</u>r ( $\frac{miles}{hour}$ ) so, this means that we have to know how many miles Izumi ran, given that the problem only mentioned<u> yards (110 yards).</u>

Let's convert 110 yards to miles, provided that he Izumi ran 1,760 yards in a mile.

110 yards ÷ 1760 $\frac{yards}{mile}$ = 0.0625 miles (this is the distance covered by Izumi in miles)

Let's go back again to the formula: s = $\frac{d}{t}$

s = $\frac{0.0625 miles}{16 seconds}$ = 0.0039 $\frac{miles}{sec}$

Since, the we arrived at a miles per second unit, we have to convert it to miles per hour.

So, if a minute has 60 seconds, then an hour has 3,600 seconds.

Thus, 0.0039 $\frac{miles}{sec}$ × 3,600 $\frac{seconds}{hour}$ = 14.04 miles per hour (the answer)

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Amy scheduled herself from 8am-1pm sunday through wednesday on wednesday she needed to release her intervals from 11am-1pm at th

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Answer:

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Step-by-step explanation:

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1 year ago

A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y de

strojnjashka [21]

Answer:

a) P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

a= μ-3.16*σ , b= μ+3.16*σ

b) P(Y≥ μ+3*σ ) ≥ 0.90

b= μ+3*σ

Step-by-step explanation:

from Chebyshev's inequality for Y

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

where

Y = the number of fish that need be caught to obtain at least one of each type

μ = expected value of Y

σ = standard deviation of Y

P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean

k= parameter

thus for

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

P{a≤Y≤b} ≥ 0.90 → 1-1/k² = 0.90 → k = 3.16

then

P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

using one-sided Chebyshev inequality (Cantelli's inequality)

P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)

P{Y≥b} ≥ 0.90 → 1- σ²/(σ²+λ²)= 1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ

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2 years ago

Which products result in a difference of squares? Check all that apply. (5z + 3)(–5z – 3) (w – 2.5)(w + 2.5) (8g + 1)(8g + 1) (–

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Answer:

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Step-by-step explanation:

* Lets explain what is the a difference of two squares

- If we multiply two binomial and the answer just two terms with

negative sign between them and the two terms are square numbers

we called this answer a difference of two squares

- Examples

# (a + b)(a - b)

- Lets multiply them

∵ (a × a) + (a × -b) + (b × a) + (b × -b)

∴ a² - ab + ba - b²

- Add the like term

∵ ab = ba

∴ -ab + ba = 0

∴ (a + b)(a - b) = a² - b² ⇒ difference of two squares

- From above the difference of two squares appears when we

multiply sum and difference of the same two terms

# (a + b) ⇒ is the sum of a and b

# (a - b) ⇒ is the difference of a and b

* Now lets solve the problem

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∵ (5z + 3) ⇒ is the sum of 5z and 3

∵ (-5z - 3) ⇒ is the difference of -5z and 3

∵ 5z ≠ - 5z

∴ They are not the sum and difference of the same two terms

∴ The product result not in a difference of squares

- In (w - 2.5)(w + 2.5)

∵ (w - 2.5) is the difference between w and 2.5

∴ (w + 2.5) is the sum of w and 2.5

∴ They are the sum and difference of the same two terms

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- In (8g + 1)(8g + 1)

∵ The two brackets are the sum of 8g and 1

∴ They are not the sum and difference of the same two terms

∴ The product result not in a difference of squares

- In (-4v - 9)(-4v + 9)

∵ (-4v - 9) is the difference between -4v and 9

∵ (-4v + 9) is the sum of -4v and 9

∴ They are the sum and difference of the same two terms

∴ The product result in a difference of squares

- In (6y + 7)(7y - 6)

∵ (6y + 7) is the sum of 6y and 7

∵ (7y - 6) is the difference between 7y and 6

∵ 6y ≠ 7y and 7 ≠ 6

∴ They are not the sum and difference of the same two terms

∴ The product result not in a difference of squares

- In (p - 5)(p - 5)

∵ The two brackets are the difference of p and 5

∴ They are not the sum and difference of the same two terms

∴ The product result not in a difference of squares

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