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Lemur [1.5K]
3 years ago
11

an art teacher has a roll of mural paper that is 5 meters long. she needs to cut it into 1- decimeter long pieces for a collage

project. how many 1- decimeter pieces can she cut from the roll of paper?
Mathematics
1 answer:
lutik1710 [3]3 years ago
7 0
50 1-decimeter pieces
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Which value of the 9 ten times in the number 920
umka21 [38]

Answer:

9000. the value of 9 in 920 is 900. and ten times anything is basically just adding a 0. so ten times 900 is 9000.

Step-by-step explanation:

Please mark brainliest and have a great day!

3 0
3 years ago
PLZ HELP I WILL MARK AS BRAINLIEST
-Dominant- [34]

Answer:

  7 square units

Step-by-step explanation:

As with many geometry problems, there are several ways you can work this.

Label the lower left  and lower right vertices of the rectangle points W and E, respectively. You can subtract the areas of triangles WSR and EQR from the area of trapezoid WSQE to find the area of triangle QRS.

The applicable formulas are ...

  area of a trapezoid: A = (1/2)(b1 +b2)h

  area of a triangle: A = (1/2)bh

So, our areas are ...

  AQRS = AWSQE - AWSR - AEQR

  = (1/2)(WS +EQ)WE -(1/2)(WS)(WR) -(1/2)(EQ)(ER)

Factoring out 1/2, we have ...

  = (1/2)((2+5)·4 -2·2 -5·2)

  = (1/2)(28 -4 -10) = 7 . . . . square units

4 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
Your class hopes to collect at least 325 cans of food for the annual food drive. there were 135 cans donated the first week and
vampirchik [111]

They have collected 135+89=224 cans until now. To equal 325, they need 325-224=101 cans more.

A.

Let C be the number needed to equate atleast 325.

135+89+C\geq 325


B.

If we solve the inequality in part A, we can find minimum number of C to meet the goal of 325 cans.

135+89+C\geq 325\\224+C\geq325\\C\geq325-224\\C\geq101

<em>So they need minimum 101 to complete the target and more than that to surpass.</em>

ANSWER: 101 cans are needed to meet the goal and more than that to surpass the goal.

5 0
3 years ago
Evaluate the sum of the finite geometric series.<br> 1+2+4+8+...+128
Wittaler [7]
S = a*((1 - r^(n+1))/(1-r)) S = 1*((1 - 2^(7+1))/(1-2)) S = 1*((1 - 2^8)/(1-2)) S = 1*((1 - 256)/(1-2)) S = 1*((-255)/(-1)) S = 255 So 1+2+4+8+...+128 = 255
7 0
3 years ago
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