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3 years ago

I need help with 1-10 Answer as many as you can please and Thank you !

Mathematics

1 answer:

Vaselesa [24]3 years ago

8 0

1. one hour and 25 minutes

2. 1,007.993

3.Yes it can. How? Simple. Can a triangle tessellate. Yes Inference in to a 3-D shape the same exact result.

4. 9:00 AM

5. 20 x 3 = 60

6. 4 x 8 = 32

7. 20 - 13 = 7

this is all i know i hope this helps :0)

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What is the value of P?

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Answer:

5P= 2P+21+45 ( sum of two sides is is equal to

the opposite exterior angle)

5P=2P +61

5P-2P=61

3P=61

p= 61÷3

p=2.33333

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3 years ago

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Triangle DEF, with vertices D(2, 5), E(6, 4), and F(3, 3), is reflected across the line y=x. A student determined one of the ver

4vir4ik [10]

The vertices of the image reflected over the line y = x are:
D´( 5 , 2 ), E´( 4, 6 ) and F´( 3, 3 ).
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3 years ago

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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

I need help with this if you get it right I will mark you brainlest

laiz [17]

The correct answer is A!

A. 1/2 is equal to .5 so multiply 2,000 by .5 and you’ll get $1000. She spends $900 on rent. Since $900 is less than $1000 she is not spending more than half of her income on rent. There for this statement is False

B. If you do 2,000 multipled by .15 (which equals 15%) then you’ll get $300 and that’s how much she spends on savings. So this statement is true.

C. 1/4 = .25 so if you do 2,000 multiplied by .25 you’ll end up with $500. The total cost of utilities, cable and groceries ( 120 + 80 + 320 ) = $520 and since $500 is less than $520 then she is spending more than 1/4 of her income on those expenses. Which makes this statement true.

D. 14% is = .14 so if you do 2,000 multiplied by .14 you’ll get $280. The total of cell phone and other expenses ( 100 + 180 ) = $280 so that is true.

8 0

3 years ago

PLEASE I NEED HELP PLEASE

mr_godi [17]

Answer:

none

Step-by-step explanation:

there's only ten cubes

3 0

2 years ago