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3 years ago

11

A ball is drawn at random from a box containing 6 red balls,4 white balls and 5 blue balls. what is the probability of picking a

blue balls

1 answer:

lions [1.4K]3 years ago

7 0

Answer:

1/3

Step-by-step explanation:

6+4+5=15 total number of balls

blue = 5 balls

5/total number = 5/15

=1/3

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<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B%282x%2B1%29%28x-5%29%7D%7B%28x-5%29%28x%2B4%29%5E%7B2%7D%20%7D" id

dybincka [34]

i) The given function is

$f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}$

The domain is

$(x-5)(x+4)^2\ne 0$

$(x-5)\ne0,(x+4)^2\ne 0$

$x\ne5,x\ne -4$

ii) For vertical asymptotes, we simplify the function to get;

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

The vertical asymptote occurs at

$(x+4)^2=0$

$x=-4$

iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

$2x+1=0$

$2x=-1$

$x=-\frac{1}{2}$

iv) To find the y-intercept, we substitute $x=0$ into the reduced fraction.

$f(0)=\frac{(2(0)+1)}{(0+4)^2}$

$f(0)=\frac{(1)}{(4)^2}$

$f(0)=\frac{1}{16}$

v) The horizontal asymptote is given by;

$lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0$

The horizontal asymptote is $y=0$ .

vi) The function has a hole at $x-5=0$ .

Thus at $x=5$ .

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

6 0

3 years ago

Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%

Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

$\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7$

$s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702$

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

$t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262$

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}$

$=176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)$

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0

3 years ago

PLEASE HELP WILL GIVE 10 POINTS!<br><br> Subtract. −3 2/7−(−5/7)

SashulF [63]

your answer would be -2.57142857143

6 0

3 years ago

A figure can be covered by 28 unit squares,without any gaps or overlaps.what is the area of the figure?

Tom [10]

Remember that the area of a square is just one of its sides squared. $A=s^2$
where
$A$ is the area of the square
$s$ is one of the sides of the square

Unit square is a square whose sides have length 1, so lets use our formula to find the volume of one unit square:
$A=s^2$
$A=1^2$
$A=1$
We now know that each one of the squares has area 1 unit squared. Since there are 28 unit square in our figure, we are going to multiply the area of a unit square by 28 to find the area of our figure:
Area of the figure= $28(1)=28$ units squared

We can conclude that the area of the figure is 28 units squared.

7 0

3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago