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natka813 [3]
3 years ago
11

All of the following represent the same ratio except _____. 1 to 5 5/1 5:1

Mathematics
1 answer:
natita [175]3 years ago
7 0

Answer:

1 to 5

Step-by-step explanation:

Others are saying 5 to 1

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Through: (4, -3), perp. to y=-2x + 1
larisa86 [58]

Step-by-step explanation:

y = -2x + 1. => Slope = -2

Then the perpendicular line will have a slope of -1/(-2) = 0.5.

We have y = 0.5x + c.

When x = 4, y = -3.

=> (-3) = 0.5(4) + c, c = -5.

Hence the answer is y = 0.5x - 5. (C)

7 0
2 years ago
Bob is having a party and wants to order some cakes.
Dmitry [639]

Answer: There isn't a possible way you can really answer this question because I don't know how much one piece cost in the first place , but if I had to guess I would say around 50-80 dollars for a cake is enough for 57 guests.

explanation: I say this because I know usually cakes for a decent amount of people are around 50-80 dollars.

7 0
3 years ago
PLS HELP HURRY ASP BEST ANSWER GETS BRAINLESST!​
a_sh-v [17]

Answer:

Hi there!

The following answers:

3. 7 c > 3 (1/4) pt   4. 5 gal > 18 qt

5. 45 cm = 450 mm 6. 4.5 km < 5000 m

Step-by-step explanation:

There are 2 cups in one pint.

There are 4 quarts in one gallon.

One kilo meter is 1000 meters.

3 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
Marika and her three friends attend the school play. Tickets cost 5.75 each, and marika paid for everyone. Find the total cost o
gavmur [86]
In all, 4 people.

5.75 for each person would be the equation

4*5.75

That equals 23.
Marina payed $23
3 0
3 years ago
Read 2 more answers
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