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2 years ago

Could someone quickly sum up how to calculate probability? I’m taking a course on it and don’t understand.

Mathematics

2 answers:

masya89 [10]2 years ago

7 0

Divide the number of events by the number of possible outcomes

Valentin [98]2 years ago

7 0

Answer:

Divide the number of events by the number of possible outcomes. After determining the probability event and its corresponding outcomes, divide the total number of ways the event can occur by the total number of possible outcomes. For instance, rolling a die once and landing on a three can be considered one event.

Step-by-step explanation:

for more information check the above attachment

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How many decimal places are in the product of the expression below?

jarptica [38.1K]

There are Two decimals

8 0

3 years ago

Read 2 more answers

Please help! Please!!

Elza [17]

You can proceed like this,

650 kb for a pic
and 2200kb for song ( 650 + 1550)

total= 18150 kb and 16 files

so, 650 kb each in both pic and song = 16 * 650= 10400

remaining = 18150-10400 = 7750 , which is contributed by additional 1550 of songs that is
7750/1550 = 5

so, a total of 5 songs and 11 pictures were sold..

hope it helped

8 0

3 years ago

if a company has 5 employees with annual saleries of $40,000, $50,000, $40,000, $60,000 and $90,000 what is the mean anulal sala

krek1111 [17]

To find the mean of something, add all the numbers then divide by the number of numbers there are.
40,000 + 50,000 + 40,000 + 60,000 + 90,000 = 280,000
Then divide:
280,000/5 = 56,000

The mean of the annual salaries at the company is 56,000

Hope this helped! If you have anymore questions or don't understand, please comment or DM me. :)

8 0

3 years ago

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed

AURORKA [14]

Answer:

A 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Step-by-step explanation:

We are given that Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured.

For the 25-mil film, the sample data result is: Mean Standard deviation 1.15 0.11 and For the 20-mil film the data yield: Mean Standard deviation 1.06 0.09.

Firstly, the pivotal quantity for finding the confidence interval for the difference in population mean is given by;

P.Q. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean speed for the 25-mil film = 1.15

$\bar X_1$ = sample mean speed for the 20-mil film = 1.06

$s_1$ = sample standard deviation for the 25-mil film = 0.11

$s_2$ = sample standard deviation for the 20-mil film = 0.09

$n_1$ = sample of 25-mil film = 8

$n_2$ = sample of 20-mil film = 8

$\mu_1$ = population mean speed for the 25-mil film

$\mu_2$ = population mean speed for the 20-mil film

Also, $s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} }$ = $\sqrt{\frac{(8-1)\times 0.11^{2}+ (8-1)\times 0.09^{2}}{8+8-2} }$ = 0.1005

Here for constructing a 98% confidence interval we have used a Two-sample t-test statistics because we don't know about population standard deviations.

So, 98% confidence interval for the difference in population means, ( $\mu_1-\mu_2$ ) is;

P(-2.624 < $t_1_4$ < 2.624) = 0.98 {As the critical value of t at 14 degrees of

freedom are -2.624 & 2.624 with P = 1%}

P(-2.624 < $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < 2.624) = 0.98

P( $-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < $2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ) = 0.98

P( $(\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ) = 0.98

98% confidence interval for ( $\mu_1-\mu_2$ ) = [ $(\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ]

= [ $(1.15-1.06)-2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ , $(1.15-1.06)+2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ ]

= [-0.042, 0.222]

Therefore, a 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Since the above interval contains 0; this means that decreasing the thickness of the film doesn't increase the speed of the film.

7 0

3 years ago

Find the volume of each three dimensional figure (3 problems please help)

grin007 [14]

Answer:

Step-by-step explanation:

1) The solid is a cube. The formula for determining the volume of a cube is expressed as

Volume = s³

s = 5 cm

Therefore,

Volume = 5³ = 125 cm³

2) The solid is a cuboid. The formula for determining the volume of a cuboid is expressed as

Volume = lwh

l = 10 inches

w = 5 inches

h = 2 inches

Therefore,

Volume = 10 × 5 × 2 = 100 inches³

3) The solid is a square base pyramid. The formula for determining the volume of a square base pyramid is expressed as

Volume = Area of square base × height

Area of square base = 3² = 9 m²

h = 10 m

Volume = 9 × 10 = 90m³

8 0

3 years ago