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2 years ago

8

. A manager has just received the expense checks for six of her employees. She randomly distributes the checks to the six employ

ees. What is the probability that exactly five of them will receive the correct checks (checks with the correct names)?

1 answer:

Tcecarenko [31]2 years ago

3 0

Answer:

13,8%

Step-by-step explanation:

There are six employees and six cheks, so there are 36 (6x6) possible combinations so if we need to measure the probability that five of them receive the exact check is only one for each one of them over the 36 possibilities, so 1/36 for one plus 1/36 the second and so on. 5/36 = 13,8%.

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Find the equation of the surface. a standard cone with vertex at (6, 0 , 0) opening up on the positive x-axis.

DerKrebs [107]

Solution: Since, it’s a standard cone opening up on the positive x- axis

Therefore, x=√y2+z2

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2 years ago

What is log Subscript 15 Baseline 2 cubed rewritten using the power property?

nalin [4]

Answer:

The equivalent expression

Step-by-step explanation:

Assuming your expression is $\log_{15}2^3$ , then we can use the power property of logarithm , which is $\log_ab^n=n\log_ab$ .

If we let a=15, b=2, and n=3

Then using the power property of logarithm will give us:

$\log_{15}2^3=3\log_{15}2$

Using the power property of logarithm, the equivalent expression is $3\log_{15}2$

7 0

2 years ago

Read 2 more answers

f(x) = −16x2 + 24x + 16 Part A: What are the x-intercepts of the graph of f(x)? Show your work. (2 points) Part B: Is the vertex

DerKrebs [107]

Answer:

see below

Step-by-step explanation:

f(x) = −16x^2 + 24x + 16

Set equal to zero to find the x intercepts

0 = −16x^2 + 24x + 16

Factor out -8

0 = -8(2x^2 -3x-2)

Factor

0 = -8(2x +1) (x-2)

Using the zero product property

2x+1 =0 x-2 =0

x = -1/2 x=2

The x intercepts are -1/2 ,2

Since the coefficient of x^2 is negative the graph will open down and the vertex will be a maximum

The x value of the maximum is 1/2 way between the zeros

(-1/2+2) /2 = 1.5/2 =.75

To find the y value substitute into the function

f(.75) = -8(2x +1) (x-2)

=-8(2*.75+1) (.75-2)

= -8(2.5)(-1.25)

=25

The vertex is at (.75, 25)

We have the zeros, and the vertex. We know the graph is symmetrical about the vertex

3 0

2 years ago

Amee received an end of year bonus of $1550 at work and went on a shopping spree. She spent $225 at the department store, $275 a

atroni [7]

Answer: She has $1022 left.

Step-by-step explanation:

The total amount that Amee received an the end of year as bonus at work is $1550.

She went on a shopping spree, spending $225 at the department store, $275 at the home furnishing store, and $28 at the card shop. Therefore, the total amount that she spent at the department store, the home furnishing store, and the card shop is

225 + 275 + 28 = $528

Therefore, the total amount of her bonus that she has left is

1550 - 528 = $1022

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2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago