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kramer
3 years ago
10

I need help fast plss

Mathematics
1 answer:
nalin [4]3 years ago
4 0

Answer:

F = FALSE

T = TRUE

A-F

B-T

C-T

D-F

E-T

F-F

G-F

Step-by-step explanation:

I'm not sure how to explain this but I'm sure someone will help, sorry, have a great day

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Mrs Larson bought a baby gift for her niece that cost 46.65 if the tax rate was 9.6% how much did she pay in tax
Ksju [112]

Answer:

2.1459

Step-by-step expnation                                                                                                        

8 0
4 years ago
What does 1,000,000,000divided by 18 • 20 - 108 equal?
Ludmilka [50]

Answer:

3,968,253.968253968

Step-by-step explanation:

7 0
3 years ago
Log(-2x+9)= log(7-4x)
expeople1 [14]

Answer:

x = -1

Step-by-step explanation:

log(-2x+9)= log(7-4x)

becomes:

-2x + 9 = 7 - 4x   as long as these are not negative

2x + 9 = 7

2x = -2

x = -1

check

-2*(-1) + 9 = 2 + 9 = 11

ok

5 0
3 years ago
In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
A circular cloth with a diameter of 20 inches is placed on a circular tray with a diameter of 24 inches. What fraction of the tr
son4ous [18]

Answer:

the fraction of surface that is not covered is 11/36

Step-by-step explanation:

To do this we first have to calculate the two surfaces

s = π * r^2

π = 3.14

r = d/2

d1 = 20

d2 = 24

r1 = 20/2

r1 = 10

r2 = 24/2

r2 = 12

s1 = π * r1^2

s1 = 3.14 * 10^2

s1 = 3.14 * 100

s1 = 314

s2 = π * r2^2

s2 = 3.14 * 12^2

s2 = 3.14 * 144

s2 = 452.16

to know the part that is not covered we simply do the s1 over the s2 and we subtract that from 1

s1 / s2 =

314 / 452.16 =

25/36

1 we subtract this

1 - 25/36 = 11/36

the fraction of surface that is not covered is 11/36

3 0
3 years ago
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