Thanks for who helps me

What is the probability that either two heads or three heads will be thrown if six fair coins are tossed at once?

lesantik [10]

Answer:The right answer will be

0. 546875

Step-by-step explanation:

When 6 coins are tossed , the total number of outcomes is 2*2*2*2*2*2= 64

The number of ways of getting two heads are 6!/2!*4!=15.

The probability of getting two heads P(A) is 15/64 .

In the same way, the probability of getting three heads P( B) is 5/16.

Now, P(A)+P(B)= 0.546875

8 0

3 years ago

What is the volume of the right prism with height h=14 cm, if the base of the prism is a triangle ∆ABC with side AB = 9 cm and a

Lina20 [59]

Answer:
volume of prism = 378 cm³

Explanation:
The volume of the right prism can be calculated as follows:
volume of prism = area of base * height of prism

1- getting the area of base:
Our base is a triangle, this means that:
area = 0.5 * base * height
We have:
base = 9 cm
height = 6 cm
Therefore:
area of base = 0.5 * 9 * 6
area of base = 27 cm²

2- getting the volume:
volume of prism = area of base * height of prism
volume of prism = 27 * 14
volume of prism = 378 cm³

Hope this helps :)

7 0

3 years ago

Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad battery life. Batt

BartSMP [9]

Answer:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

$H_o$ : $\sigma _1 ^2 \leq \sigma _2^2$

Alternative hypothesis: the variance in hours of usage for talking is greater than the the variance in hours of usage for internet.

$H_a : \sigma_1^2 > \sigma_2^2$

$\mathbf{s_ 1 =16.11}$

$\mathbf{s_2 = 7.98}$

Step-by-step explanation:

Let $x_1$ and $x_2$ be the two variables that represents the battery life in hours for talking usage and battery life in hours for internet usage respectively.

The hypothesis can be formulated as:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

$H_o$ : $\sigma _1 ^2 \leq \sigma _2^2$

Alternative hypothesis: the variance in hours of usage for talking is greater than the the variance in hours of usage for internet.

$H_a : \sigma_1^2 > \sigma_2^2$

The standard deviation for the battery usage for talking is :

$\bar x_1 = \dfrac{1}{n_1} \sum x_i \\ \\ \bar x_1 = \dfrac{1}{12}(35.8 +22.4+...+35.5) \\ \\ \bar x_1 = \dfrac{241.2}{12} \\ \\ \bar x_1 =20.1$

The standard deviation Is:

$s_ 1 = \sqrt{\dfrac{1}{n_1-1}\sum (x{_1i}-\bar x_i)^2}$

$s_ 1 = \sqrt{\dfrac{1}{12-1}\sum (35.8- 20.1)^2+ (35.5-20.1)^2}$

$s_ 1 = \sqrt{259.568}$

$\mathbf{s_ 1 =16.11}$

The standard deviation for the battery life usage for the internet is :

$\bar x_2 = \dfrac{1}{n_2} \sum x_{2i}$

$\bar x_2 = \dfrac{1}{10} (24.0+12.5+36.4+...+4.7})$

$\bar x_2 = \dfrac{115}{10}$

$\bar x_2 = 11.5$

Thus; the standard deviation is:

$s_2 = \sqrt{\dfrac{1}{n_2-1}(x_{2i}- \bar x_2)^2}$

$s_2 = \sqrt{\dfrac{1}{10-1}(24-11.5)^2+(4.7-11.5)^2}$

$s_2 = \sqrt{63.60}$

$\mathbf{s_2 = 7.98}$

4 0

3 years ago

Which is greater 1/3 or 30%

Brilliant_brown [7]

30% = 30/100 = (30 x 3)/(100 x 3) = 90/300

1/3 = (1 x 100)/(3 x 100) = 100/300

As 90/300 < 100/300

Therefore 30% < 1/3

4 0

4 years ago

Two buildings on opposites sides of a highway are 3x^3- x^2 + 7x +100 feet apart. One building is 2x^2 + 7x feet from the highwa

iVinArrow [24]

Given:

Distance between two buildings = $3x^3- x^2 + 7x +100$ feet apart.

Distance between highway and one building = $2x^2 + 7x$ feet.

Distance between highway and second building = $x^3 + 2x^2 - 18$ feet.

To find:

The standard form of the polynomial representing the width of the highway between the two building.

Solution:

We know that,

Width of the highway = Distance between two buildings - Distance of both buildings from highway.

Using the above formula, we get the polynomial for width (W) of the highway.

$W=3x^3- x^2 + 7x +100-(2x^2 + 7x)-(x^3 + 2x^2 - 18)$

$W=3x^3- x^2 + 7x +100-2x^2-7x-x^3 -2x^2+18$

Combining like terms, we get

$W=(3x^3-x^3)+(- x^2 -2x^2-2x^2)+ (7x -7x)+(100 +18)$

$W=2x^3-5x^2+0+118$

$W=2x^3-5x^2+118$

Therefore, the width point highway is $2x^3-5x^2+118$ .

8 0

3 years ago