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3 years ago

27.36 divided by 3.14 use compatible numbers to estimate the quotient

Mathematics

2 answers:

stiks02 [169]3 years ago

6 0

Answer:

Step-by-step explanation:

you can round the numbers

27/3= 9

27.36 rounds down to 27 and 3.14 rounds to 3

ElenaW [278]3 years ago

6 0

I think 27 and 3 would be your answer hope this helps! Good luck

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A box contains three plain pencils and 5 pens. A second box contains three colored pencils and three crayons. One item from each

worty [1.4K]

Answer:

Therefore the probability that a pen from the first box and a crayon from the second box are selected is $\frac5{16}$

Step-by-step explanation:

Probability:

The ratio of the number of favorable outcomes to the number all possible outcomes of the event.

$Probability=\frac{\textrm{The number favorable outcomes}}{\textrm{The number all possible}}$

Given that,

Three plain pencils and 5 pens are contained by the first box.

Total number of pens and pencils is =(3+5)=8

The probability that a pen is selected from the first box is

=P(A)

$=\frac{\textrm{The number pens}}{\textrm{Total number of pens and pencils}}$

$=\frac{5}{8}$

A second box contains three colored pencils and three crayons.

Total number of pencils and crayons is =(3+3)=6

The probability that a crayon is selected from the second box is

=P(B)

$=\frac{\textrm{The number of crayon}}{\textrm{Total number of crayons and pencils}}$

$=\frac{3}{6}$

Since both events are mutually independent.

The required probability is multiple of the events

Therefore the required probability is

$=\frac58\times \frac36$

$=\frac5{16}$

4 0

3 years ago

What equation represents (0,3) (1, 1)

weqwewe [10]

your answer would be 2x - y + 3 = 0

3 0

2 years ago

One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R

marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

$Urn\ 1 = \{B_1, R_1, R_2, R_3\}$

$Urn\ 2 = \{R_4, R_5, B_2, B_3\}$

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

If urn 2 is selected, the following selection exists:

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

See attachment for possibility tree

Solving (b): The total number of outcome

For urn 1

There are 4 balls in urn 1

$n = \{B_1,R_1,R_2,R_3\}$

Each of the balls has 3 subsets. i.e.

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

So, the selection is:

$Urn\ 1 = 4 * 3$

$Urn\ 1 = 12$

For urn 2

There are 4 balls in urn 2

$n = \{B_2,B_3,R_4,R_5\}$

Each of the balls has 3 subsets. i.e.

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

So, the selection is:

$Urn\ 2 = 4 * 3$

$Urn\ 2 = 12$

Total number of outcomes is:

$Total = Urn\ 1 + Urn\ 2$

$Total = 12 + 12$

$Total = 24$

5 0

2 years ago

A radar gun was used to record the speed of a swimmer (in meters per second) during selected times in the first 2 seconds of a r

Natalka [10]

Trapezoidal is involving averageing the heights
the 4 intervals are
[0,4] and [4,7.2] and [7.2,8.6] and [8.6,9]

the area of each trapezoid is (v(t1)+v(t2))/2 times width

for the first interval
the average between 0 and 0.4 is 0.2
the width is 4
4(0.2)=0.8

2nd
average between 0.4 and 1 is 0.7
width is 3.2
3.2 times 0.7=2.24

3rd
average betwen 1.0 and 1.5 is 1.25
width is 1.4
1.4 times 1.25=1.75

4th
average betwen 1.5 and 2 is 1.75
width is 0.4
0.4 times 1.74=0.7

add them all up
0.8+2.24+1.75+0.7=5.49

5.49
t=time
v(t)=speed
so the area under the curve is distance
covered 5.49 meters

8 0

2 years ago

Please can someone help me with this question.

saw5 [17]

Answer:

Step-by-step explanation:

8 0

2 years ago