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svp [43]
2 years ago
6

Distributive property

Mathematics
2 answers:
Eduardwww [97]2 years ago
6 0
 Distributive property allows you multiply a sum by multiplying each addend individually and then add the products.
here is an example of distributive property: 1 (2+3) = 1 x 2 + 1 x 3 

I hope this helps!
Olenka [21]2 years ago
4 0
A(b+c) = ab+ac
just essentially expanding
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Two percent of all seniors in a class of 50 have scored above 96% on an ext exam, which of the following is the number of senior
dalvyx [7]

Answer:

The number of seniors who scored above 96% is 1.

Step-by-step explanation:

Consider the provided information.

Two percent of all seniors in a class of 50 have scored above 96% on an ext exam.

Now we need to find the number of seniors who scored above 96%

For this we need to find the two percent of 50.

2% of 50 can be calculated as:

\frac{2}{100}\times50

\frac{100}{100}

1

Hence, the number of seniors who scored above 96% is 1.

6 0
3 years ago
2 horizontal and parallel lines are intersected by 2 diagonal lines to form a triangle with exterior angles. The top angle of th
liraira [26]

Answer:

11.25

Step-by-step explanation:

sketch a triangle and fill in the sides,add the sides and equate to 180, collect the like terms to get the value of x=11.25

8 0
3 years ago
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How would I simplify number 3) ?
Anna [14]
You wouldn’t because 3 is the base number and cannot be simplified anymore
6 0
3 years ago
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A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)
Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

36^6

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

5\cdot 36^5

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

P(A)=P(B)=\dfrac{5}{36}

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8

exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8

ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8

oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8

uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}

Finally, you can compute the probability of the union using the formula

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Since we already computed all these quantities.

7 0
2 years ago
Do we ever use numbers to describe the values of a categorical variable?
Vika [28.1K]

Answer:

Yes

Step-by-step explanation:

Because A categorical variable can bexpressed as a number when statistical evaluation is intended, However, these numbers do not mean the same as a numerical value e.g giving numbers to waist sizes, test scores, grade level etc

4 0
3 years ago
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