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Marta_Voda [28]
3 years ago
6

2 1/4 x2 1/4 in mixed number

Mathematics
1 answer:
maks197457 [2]3 years ago
4 0

Answer:

5 \frac{1}{16} is the answer.

Step-by-step explanation:

Given:

2 \frac{1}{4} \times 2 \frac{1}{4}

<u>Step 1: Converting the mixed fraction into  fraction:</u>

Multiply  4 by 2 add  1 to the result.

<u></u>2\frac{1}{4}  =  \frac{4 \times 2+1}{4}

=\frac{8+1}{4}

=\frac{9}{4}

<u>Step 2: Performing the multiplication using the fraction</u>

= >\frac{9}{4} \times \frac{9}{4}

=> \frac{81}{16}

<u>Step 3: Converting the product back to  mixed fraction </u>

divide 81 by 16 and find the remainder

=>quotient \frac{remainder}{divisor}

Dividing 81 by 16 we get  5 as quotient and 1 as remainder

Thus the result in  mixed fraction will be

=> 5 \frac{1}{16}

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Which two number have an absolute value of 5
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Answer:

5 and negative 5

Step-by-step explanation:

The absolute value is the distance from 0 five and negative 5 are both the same distance from zero

7 0
2 years ago
The pool holds 17,922 gallons of water, and is leaking at a rate of 4 gallons per day. If Mike does not replace the water that h
emmasim [6.3K]

Answer:

17442

Step-by-step explanation:

120x4= 480

17922-480=

3 0
3 years ago
Read 2 more answers
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
Solve the given inequality 11x-20&lt;_13​
RUDIKE [14]

Answer:

x ≤ 3

Step-by-step explanation:

11

x

−

20

≤

13

Step 1: Add 20 to both sides.

11

x

−

20

+

20

≤

13

+

20

11

x

≤

33

Step 2: Divide both sides by 11.

11

x

11

≤

33

11

3 0
2 years ago
Read 2 more answers
A soup can and label are shown. The label wraps around the circumference of the cylindrical can.
den301095 [7]

Answer: i would say 60 cm sorry if im wrong i think your missing a number

Step-by-step explanation:

4 0
3 years ago
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