Ask question

Ask question

All categories

3 years ago

15

Each letter from the word MATHEMATICS is written on a separate slip of paper. The 11th slips of paper are placed in a hat and tw

o slips are drawn at random. If the first letter is not replaced find the probability that the first letter is a vowel (a,e,I,o,u) and the second letter is a vowel

1 answer:

amid [387]3 years ago

6 0

It may be possibly 4 out of 9

You might be interested in

F(x) = 9x^2+30x+25 pls help

lapo4ka [179]

Answer: f(x) =(3x+5)^2

x= -5/3

Step-by-step explanation:

5 0

3 years ago

Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t

Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1. $f(x)=\frac{x^2+x-20}{x^2+4}$

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345$

3. $h(x)=\frac{3x-5}{x^2-5x+7}$

$x^2-5x+7=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

$\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43$

2. $g(x)=\frac{x-17}{x^2+75}$

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12$

4. $i(x)=\frac{x^2-9}{x-9}$

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is not continuous on the set of real numbers.

5. $j(x)=\frac{4x^2-7x-65}{x^2+10}$

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0$

6. $k(x)=\frac{x+1}{x^2+x+29}$

$x^2+x+29=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

$\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102$

7. $l(x)=\frac{5x-1}{x^2-9x+8}$

$x^2-9x+8=0$

$x^2-8x-x+8=0$

$x(x-8)-1(x-8)=0$

$(x-8)(x-1)=0$

$x=8,1$

The function is not defined for x=8 and x=1

Hence, function l is not defined for all real values.

8. $m(x)=\frac{x^2+5x-24}{x^2+11}$

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722$

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

6 0

3 years ago

2. Jessica is a waitress. Her customer’s bills totaled $300. She earned $45 in tips. What percent did her customers tip her? ___

zepelin [54]

Answer:

15%

Step-by-step explanation:

$45/$300 = 0.15 = 15%

4 0

3 years ago

Read 2 more answers

The experimental factor that is manipulated; the variable whose effect is being studied is called ____________

lorasvet [3.4K]

Answer:

The experimental factor that is manipulated; the variable whose effect is being studied is called <u>independent variable.</u>

Step-by-step explanation:

Consider the provided information.

In an experiment, the two principal variables are the independent and dependent variable.

An independent variable is the variable that is altered or controlled to test the effects on the dependent variable in a scientific experiment.

The variable which is tested and measured in a scientific experiment is a dependent variable.

From the above definition: The experimental factor that changed or controlled in a scientific experiment is called independent variable.

Therefore, the complete statement is: The experimental factor that is manipulated; the variable whose effect is being studied is called <u>independent variable.</u>

8 0

3 years ago

Find the value of this expression if x=-1 and y=-5 x^2y/4

REY [17]

Answer:

$\large\boxed{-\dfrac{5}{4}=-1\dfrac{1}{4}}$

Step-by-step explanation:

$\text{Put the values of x = -1 and y = -5 to the expression}\ \dfrac{x^2y}{4}:\\\\\dfrac{(-1)^2(-5)}{4}=\dfrac{(1)(-5)}{4}=\dfrac{-5}{4}$

4 0

3 years ago