What is the runner’s average rate of change between the hours: 0.5 and 2? mph 1.5 and 2.5? mph Average Rate of Change

a. at r = 2

P = [10! / (10 – 2)! 2!] * 0.33^2 * 0.67^(10 – 2)

P = 0.1989 ~ 0.20               (answer)

b. We deduct the values of P when r = 0, r = 1, r = 2 from 1 to get all values when r > 2

at r = 0

P = [10! / (10 – 0)! 0!] * 0.33^0 * 0.67^(10 – 0)

P = 0.0182

at r = 1

P = [10! / (10 – 1)! 1!] * 0.33^1 * 0.67^(10 – 1)

P = 0.0898

so for r > 2, the probability is:

P = 1 – (0.1989 + 0.0182 + 0.0898)

P = 0.6931            (answer)

c. at r = 3

P = [10! / (10 – 3)! 3!] * 0.33^3 * 0.67^(10 – 3)

P = 0.2614

at r = 4

P = [10! / (10 – 4)! 4!] * 0.33^4 * 0.67^(10 – 4)

P = 0.2253

at r = 5

P = [10! / (10 – 5)! 5!] * 0.33^5 * 0.67^(10 – 5)

P = 0.1332

So when r is between 2 and 5 inclusive, the probability is:

P = 0.1989+ 0.2614 + 0.2253 + 0.1332

P = 0.8188            (answer)

The bank's likely response to Kevin's concern would be that checks are recorded on the date they are cashed, not on the date they are written, which explains the mismatch in dates Kevin observed. Balancing one's checkbook regularly is important for managing personal finances.

When Kevin has noticed that the dates on his check register do not match the dates on his account statement and contacts the bank, the bank's likely response will be "The date that the check is written is usually a few days before it is cashed. We record when it was cashed and you record when it's written." This is a common situation in personal finance and is related to the cash basis of accounting that most individuals use to manage their finances. The difference in dates arises because checks are processed and cleared by the bank on the date they are cashed, not on the date they are written. It's important for individuals to balance their checkbooks regularly to manage money, keep track of cash flow, and avoid overdraft fees.

Answer:

The station first give away all three prizes to 300th caller.

Step-by-step explanation:

A radio station gives away $15 to every 15th caller, $25 to every 25th caller, and free concert tickets to every 100th caller.

To get the answer we need to calculate L.C.M.  of 15, 25, 100.

15 = 3 x 5

25 = 5 x 5

100 = 2 x 2 x 5 x 5

LCM is = 2 x 2 x 3 x 5 x 5 = 300.

Hence, the station first give away all three prizes to 300th caller.

Answer:

[tex]i^{12}=1[/tex]

Last option is correct.

Step-by-step explanation:

We have to simplify the expression [tex]i^{12}[/tex]

We can rewrite this expression using the properties [tex](a^m)^n=a^{mn}[/tex]

[tex](i^4)^{3}[/tex]

Now, we know that [tex]i^4=1[/tex]

Hence, we have

[tex](1)^{3}[/tex]

On simplifying, we get

[tex]1[/tex]

Thus, the simplified form is 1.

Last option is correct.

Answer:

it's 1

Step-by-step explanation:

Answer:

All the numbers are less than and equal to 18.

Step-by-step explanation:

To find : All numbers such that a third of a number increased by half that number is at least 3 less than that same number ?

Solution :

Let the number be 'x',

A third of a number is [tex]\frac{x}{3}[/tex]

Half that number is [tex]\frac{x}{2}[/tex]

3 less than that same number is [tex]x-3[/tex]

A third of a number increased by half that number is at least 3 less than that same number is written as,

[tex]\frac{x}{3}+\frac{x}{2}\geq x-3[/tex]

[tex]\frac{2x+3x}{6}\geq x-3[/tex]

[tex]\frac{5x}{6}\geq x-3[/tex]

[tex]5x\geq 6x-18[/tex]

[tex]5x-6x\geq -18[/tex]

[tex]-x\geq -18[/tex]

[tex]x\leq 18[/tex]

Therefore, all the numbers are less than and equal to 18.

The set of numbers such that a third of a number increased by half that number is at least 3 less than that same number are given as; x <= 18.

According to the question;

Let the numbers in question be represented as x.

In essence;

[tex] \frac{x}{3} + \frac{x}{2 } \geqslant x - 3[/tex]

[tex] \frac{5x}{6} \geqslant x - 3[/tex]

Therefore, the set of numbers such that a third of a number increased by half that number is at least 3 less than that same number are given as; x <= 18.

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To find the law connecting w and t, we establish w as partly constant and partly inverse to t squared. Using given values, we find constants k and c and then solve for t when w is 46.

The problem states that w is partly constant and partly varies inversely as the square of t. Given two sets of values for w and t, we can establish the relationship between them. Since the variation is inverse with respect to the square of t, we can write the relationship as:

w = k / t2 + c

where k is a constant of variation and c is the constant part of w. To find these constants, we use the given data points (24, 4) and (18, 2). Substituting these into the equation, we get two simultaneous equations:

Solving these will give us the values of k and c, which then allows us to find t when w = 46.

To solve for t when w = 46, we plug this value into our established equation and solve for t. We'll end up with a quadratic equation in terms of t that we need to solve to find the time t.

Answer:

The greatest perfect square that is a factor of 980 is 196.

Step-by-step explanation:

To find : What is the greatest perfect square that is a factor of 980?

Solution :

First we factor the number 980.

[tex]980=2\times 2 \times 5\times 7\times 7[/tex]

[tex]980=2^2\times 7^2\times 5[/tex]

[tex]980=(2\times 7)^2\times 5[/tex]

[tex]980=(14)^2\times 5[/tex]

[tex]980=196\times 5[/tex]

This clearly means that 196 is the required factor.

Therefore, The greatest perfect square that is a factor of 980 is 196.

We have that The Greatest perfect square of 980 is

x= 196

From the question we are told that

The greatest perfect square that is a factor of 980

Generally

Greatest perfect square

Where a  perfect square is a number that is written to given the product of two identical Numbers

Therefore

Greatest perfect square of 980 is

x= 196

Where

The factors of 980 are 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 49, 70, 98, 140, 196, 245, 490, 980

Therefore

The Greatest perfect square of 980 is

x= 196

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x - 2/3 = 5/6

 the only step needed is to add 2/3 to 5/6

2/3 + 5/6 = 1 1/2

x = 1 1/2

Answer:

the value of x is [tex]\frac{3}{2}[/tex]

Step-by-step explanation:

we need to solve value of x for expression [tex]x-\frac{2}{3}=\frac{5}{6}[/tex]

[tex]x-\frac{2}{3}=\frac{5}{6}[/tex]

Add both the sides by [tex]\frac{2}{3}[/tex]

[tex]x-\frac{2}{3}+\frac{2}{3}=\frac{5}{6}+\frac{2}{3}[/tex]

take L.C.M both the side and simplify

[tex]x=\frac{3}{2}[/tex]

Hence, the value of x is [tex]\frac{3}{2}[/tex]

The value 0.02% written as a decimal is 0.0002

Given the value to be written as a decimal :

To express as a decimal , divide 0.02 by 100

Now we have ;

Therefore, the value 0.02% written as a decimal is 0.0002

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To evaluate the ratio of 390mm to 13cm, we convert 13cm to 130mm and then simplify the ratio of 390mm to 130mm to its lowest terms, which is 3 : 1.

To evaluate the ratio of 390mm to 13cm, we first need to express these measurements in the same unit. Converting centimeters to millimeters is simple: there are 10 millimeters in 1 centimeter.

Therefore, 13cm is equivalent to 130mm. Now, we can write the ratio as 390mm : 130mm. To simplify this ratio to its lowest terms, we divide both numbers by the greatest common divisor, which is 130 in this case. The simplified ratio is therefore 3 : 1.

The generalized description is as follows: quantity (in old units)  imes conversion factor = quantity (in new units). One must remember to simplify the ratio by dividing both the numerator and the denominator by their greatest common divisor, finalizing the answer without the '1' in the denominator when it's the only figure present.

The domain of the function is restricted to the altitude of the plane, and the plane cannot exceed its absolute ceiling. The graph of the function will have a vertical asymptote at the absolute ceiling. As the plane approaches its absolute ceiling, the time required to further increase its altitude increases significantly. To find the time it will take for the plane to climb to a specific altitude, substitute the altitude into the equation and solve for time.

a. The domain of the function is determined by the restrictions given in the problem. In this case, the domain represents the altitude of the plane. Since the plane's absolute ceiling is 18,000 feet, the altitude cannot exceed 18,000 feet. Therefore, the domain is 0 ≤ h ≤ 18,000.

b. To graph the function, we can plot points by substituting different values of h into the function and calculating the corresponding values of t. This will result in a smooth curve. Asymptotes occur when the function approaches certain values of h or t. In this case, the function has a vertical asymptote at h = 18,000, because the plane cannot climb beyond its absolute ceiling.

c. As the plane approaches its absolute ceiling of 18,000 feet, the time required to further increase its altitude increases significantly. This is because the logarithmic function in the equation has a slower growth rate as h approaches its limit.

d. To find the time it will take for the plane to climb to an altitude of 4000 feet, we can substitute h = 4000 into the equation t = 50 log10 (18000/(18000-h)) and solve for t.

x < -2 and y ≤ -x -2

for every 5 men there are 6 women

90 /5 = 18

18 *6 = 108 women

Answer:

108 women

Step-by-step explanation: