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tia_tia [17]
3 years ago
11

Use compatible numbers to find two estimates 18÷1322

Mathematics
1 answer:
marysya [2.9K]3 years ago
8 0
0.0135135135135135135135135 etc
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Math
denis-greek [22]

Answer:

\bullet \ \large\text{The mean in period 2 is higher than the mean in period 1.}

Step-by-step explanation:

<u><em>the mean in period</em></u> 1 :

(2.3+2.1+2.2+2.2+2.2+2.1+2.4+2.5+2.2+2.0+1.9+1.9+2.1+2.2+2.3)÷15=21.733...

<u><em>the mean in period</em></u> 2 :

(2.3+2.1+3.3+1.5+3.6+1.6+3.0+1.1+4.7+2.1+2.4+1.9+2.8+0.5+2.3)÷15=23.466...

Since 23.466 > 21.733 then “The mean in period 2 is higher than the mean in period 1”.

7 0
2 years ago
What is the length of side AC as shown on the coordinate plane? 5 4 3 NO A 00 2 3 5 X O 1 units O 3 units O 4 units Suo​
kobusy [5.1K]

the coordinates are (-2,1) (-2,-3) that is the answer

6 0
2 years ago
Given a = -3, b = 4 and c = -5, evaluate a - b - c. -12 -6 -2
KiRa [710]
A - B - C = 
-3 - 4 + 5 = 
-7 + 5 =
-2

negative three minus four plus five is negative two
welcome :/
4 0
3 years ago
Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t
LiRa [457]

Answer:

(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

Step-by-step explanation:

The sample space for the event of rolling two dice is presented below

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6)  =\dfrac{20}{36}

The probability distribution of this event is given as follows.

\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|

First, we determine the expected Value of this event.

Expected Value

=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning  =\dfrac{20}{36}

If the game is played 100 times

Number of times expected to win

=\dfrac{20}{36} \times 100\\=56$ times

Therefore, number of times expected to loose

= 100-56

=44 times

8 0
3 years ago
Answers please! I’m finding the volume and i don’t really get it
MrRa [10]

Answer:

V = 13/2

Step-by-step explanation:

When finding volume you must know the formula on doing so.

<em>Volume = length × width × height</em>

length = 2\frac{1}{2} --> \frac{5}{2}

width = 2\frac{1}{2} --> \frac{5}{2}

height = 3\frac{1}{4} --> \frac{13}{4}

                        <u>mixed number --> fraction</u>

V = \frac{5}{2} x \frac{5}{2} x \frac{13}{4}

V = \frac{25}{4} x \frac{13}{4}

V = \frac{325}{8} or \frac{13}{2}

3 0
2 years ago
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