All categories

2 years ago

What is the probability of rolling a 5 or not rolling a 5 using a regular 6-sided number cube?

2 answers:

7 0

There are 6 different sides you can land on with a dice. So the probability of rolling any one side is 1/6. The probability of NOT rolling a 5 is the same thing as rolling anything except a 5. So that probability will be 5/6.

Andrews [41]2 years ago

7 0

Well, the probability of rolling a 5 would be 1/6, the probability of not rolling a 5 would be 4/6 or 2/3.

Since the problem uses or, the question is asking for the probability of one or the other occurring, in other words the probability is more that if it were to simply be one event. Thus we add the probabilities together.

1/6 + 4/6 = 5/6.

You might be interested in

A rectangular prism has a length of 12 centimeters, a width of 6 centimeters, and a height of 9 centimeters. A cross section is

Klio2033 [76]

Answer:

<h2>The perimeter of the cross section is 30 centimeters.</h2>

Step-by-step explanation:

In this problem we have the intersection of a rectangular prism an a plane.

The dimensions of the rectangular plane are

$l=12cm\\w=6cm\\h=9cm$

Assuming the plane is cutting the prism horizontally, the cross section would have dimensions

$w=6cm\\l=9cm$

Because only the height would be cut.

So, the perimeter of the rectangle cross-section is

$P_{cross}=2(l+w)=2(6+9)=2(15)=30cm$

Therefore, the perimeter of the cross section is 30 centimeters.

3 0

2 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago

What is the slope of the line that passes through the points (-2,5) and (3,-4) in the standard (x,y) coordinate plane

Scorpion4ik [409]

To find the slope between two coordinates, you use the formula:

m = y2 - y1/x2 - x1

If coordinate A is (-2, 5) and coordinate B is (3, -4), then:

y2 = -4

y1 = 5

x2 = 3

x1 = -2

So, just substitute the points into the formula.

m = -4 - 5/3 - (-2)

m = -9/5

The slope of the line is -9/5.

3 0

2 years ago

he data represents the daily rainfall (in inches) for one month. Construct a frequency distribution beginning with a lower clas

Licemer1 [7]

Answer:

It is not normally distributed as it has it main concentration in only one side.

Step-by-step explanation:

So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).

So, let us begin the groupings into their different classes, shall we?

Data given:

0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.

(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.

(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31

(3). 0.4 - 0.6 : there are 2 values that falls into this category.

(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.

Class interval frequency.

0.00 - 0.20. 15.

0.20 - 0.40. 2.

0.4 - 0.6. 2.

4 0

2 years ago

A family of 4 spent $104 for tickets to a concert on Friday and they spent $140 for tickets to a dinner theater on Saturday. All

Julli [10]

Answer:

Friday's tickets: $26 each
Saturday's tickets: $35 each

Step-by-step explanation:

We presume that 4 tickets were bought each day, so the price of 1 ticket is 1/4 of the total price:

(1/4)($104) = $26

(1/4)($140) = $35

The cost of each ticket on Friday was $26; on Saturday, the cost was $35.

5 0

2 years ago