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3 years ago

If Teri had 5 blouses and 3 pairs of pants, how many different combinations could she wear?

2 answers:

6 0

If Terri had 5 blouses and 3 pairs of pants, Teri could wear 15 different combinations of clothes. 1 blouse can be worn with any of the three pairs of pants, as for the other 4. 5 * 3 = 15. The answer is 15 combinations.

Vitek1552 [10]3 years ago

6 0

The answer would be 15. If she has 5 and 3 you would do 5•3=15.
There is 15 combinations.

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How do I solve this?

podryga [215]

Answer:

$3078.8in^3$

Step-by-step explanation:

$V=\pi r^2h=\pi *7^2*20=3078.7608\\=3078.8in^3$

4 0

3 years ago

What is the sum of 1/3 + 5 3/4 as a mixed number in simplest form

mart [117]

6 1/12 would be the answer

Rewriting our equation with parts separated

1/3+5+3/4

Solving the fraction parts

1/3+3/4=?

Find the LCD of 1/3 and 3/4 and rewrite to solve with the equivalent fractions.

LCD = 12

4/12+9/12=13/12

Simplifying the fraction part, 13/12,

13/12=11/12

Combining the whole and fraction parts

5+1+1/12=6 1/12

3 0

3 years ago

What is the converse of the conditional statement?

Natalija [7]

Answer:

If x+1 is odd, then x is even B

Step-by-step explanation:

Conditional: If p, then q

Converse: If q, then p

p= x being even

q, x+1 being odd.

5 0

3 years ago

The probability of Roger winning the tournament, given that Stephan wins, is 0. The probability of Stephan winning the tournamen

Fofino [41]

They're both true. Both statements, together, make one
single fancy way of saying "Only one man can win.".

7 0

3 years ago

Need answer to 4 and 5

algol13

Answer:

Substitute n =1, 2, 3, 4, 5 to get the first five terms.

Step-by-step explanation:

(4). $g_n = 2 . 3^{n - 1}$

To find the first term, substitute n = 1.

Therefore, $g_1 = 2 . 3^{1 - 1} = 2 . 3^{0} = 2 . 1$ = 2

$g_2 = 2 . 3^{2 - 1} = 2 . 3$ = 6

$g_3 = 2 . 3^{3 - 1} = 2 . 3^{2} = 2 . 9$ = 18

$g_4 = 2 . 3^{4 - 1} = 2 . 3^{3} = 2 . 27$ = 54

$g_5 = 2 . 3^{5 - 1} = 2 . 3^{4} = 2 . 81$ = 168

Therefore the first five terms of the sequence are: 2, 6, 18, 54, 168.

(5). $t_n = \frac{2}{3}t_{n - 1}$

$t_2 = \frac{2}{3} t_{2 - 1} = \frac{2}{3}t_1 = \frac{2}{3}6$ = 4

$t_3 = \frac{2}{3} t_2 = \frac{2}{3}. 4 = \frac{8}{3}$

$t_4 = \frac{2}{3}t_3 = \frac{2}{3}\frac{8}{3} = \frac{16}{9}$

$t_5 = \frac{2}{3}t_4 = \frac{2}{3}\frac{16}{9} = \frac{32}{27}$

Therefore the first five terms in the sequence are: 6, 4, 8/3, 16/9, 32/27.

8 0

3 years ago