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2 years ago

6

Can someone help me on my last problem? #10

2 answers:

Mashutka [201]2 years ago

8 0

Ok but what did i do wrong and if you never studied the subject how did you know i was wrong

Kruka [31]2 years ago

8 0

Based on my small expertise in this subject,

the answer should be 1/(20)^22

the logic is the same as a coin flip
1st flip: call for heads, 1/2 chance
2nd: call for heads, (multiply by self) 1/2 x 1/2= 1/4 chance
3rd: heads, 1/4 x 1/2= 1/8
and so on...

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What are the center and radius of the circle whose equation is x2 + y2 + 4x = 5?

prisoha [69]

The center is (-2,0)
The radius is -3

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3 years ago

Jay has $8. he wants to buy a book for $7.50. the sales tax is 6%. does he have enough money? explain your reasoning.

antoniya [11.8K]

Yes he does.

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6x7.5=45

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3 years ago

Evaluate the expression 9

Sindrei [870]

Answer:

3

Step-by-step explanation:

9/3

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2 years ago

Read 2 more answers

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

5 0

2 years ago

Read 2 more answers

Chase buys a bag of cookies that contains 6 chocolate chip cookies, 6 peanut butter cookies, 6 sugar cookies and 6 oatmeal cooki

Over [174]

Answer:

0.0652173

Step-by-step explanation:

Given that :

6 chocolate chip cookies

6 peanut butter cookies

6 sugar cookies

6 oatmeal cookies

Total number of cookies purchased = (6+6+6+6) = 24

Probability, P= required outcome /total possible outcomes

This is a selection without replacement probability problem :

P(peanut butter cookies) = 6/24 = 1/4

Then ;

P(chocolate chip cookie) = 6/23

Hence,

P(peanut butter cookies then chocolate chip cookie) = 1/4 * 6/23 = 0.0652173

4 0

2 years ago