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slava [35]
3 years ago
7

Joe says that the product of a 4 digit number and a 1 digit number is always a 4 digit number does Joe's statement makes sense?

Mathematics
2 answers:
Eduardwww [97]3 years ago
5 0
No because thw product of 9 and anything above 1112 would be 10,008+
creativ13 [48]3 years ago
5 0

No, what Joe said isn't necessarily true. Though it is common that the product will be a 4 digit number, because you can use any number, like a large 4 digit number like 9999 or a large 1 digit number like 9, that isn't true. For instance: "9999 x 9" is 89991 which is 5 digits instead of 4 because "9999" is the biggest 4 digit number and "9" is the biggest "1" digit number.

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7 0
3 years ago
30 points please answer honestly. Geometry help needed. <br> D is 9.32
Vadim26 [7]

Answer:

7.75 m

Step-by-step explanation:

First find the circumference. Use the formula C = 2πr

C = 2(3.14)(3)

C = 18.84

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Multiply

18.84/1 x 148/360

2788.32/360

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4 0
3 years ago
Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find
Papessa [141]

Answer:

1) \text{P(at least one boy and one girl)}=\frac{3}{4}

2) \text{P(at least one boy and one girl)}=\frac{3}{8}

3) \text{P(at least two girls)}=\frac{1}{2}

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To  Find : The probability of each event.  

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)        

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

1) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, BGG, GBB, GBG, GGB=6

\text{P(at least one boy and one girl)}=\frac{6}{8}

\text{P(at least one boy and one girl)}=\frac{3}{4}

2) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, GBB=3

\text{P(at least one boy and one girl)}=\frac{3}{8}

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

\text{P(at least two girls)}=\frac{4}{8}

\text{P(at least two girls)}=\frac{1}{2}

4 0
3 years ago
Read 2 more answers
Regina purchased 1.75 pounds of turkey breast from her local deli for $5.99 per pound. To the nearest cent, how much did she spe
VashaNatasha [74]
5.99(1.75)= 10.4825

10.4825 to the nearest hundredth is 10.48

$10.48
3 0
3 years ago
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Answer:

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3 0
2 years ago
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