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IgorLugansk [536]
3 years ago
12

Help(show work please and thank you)

Mathematics
1 answer:
ruslelena [56]3 years ago
5 0
35/2.5=14 x=14 hope that helps
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Will give brainly and 50 points! Please be original and do it in less then 10 minutes please and ty!
stepladder [879]

Answer:

110.88 feet²

Step-by-step explanation:

Area of trapezoids = (base 1 + base 2) × (height ÷ 2)

(15.2 + 11.2) × (8.4 ÷ 2) = 110.88

The area is 110.88 feet²

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Use algebra tiles to find (6x² + 5x + 3) - (4x + 2).<br> .
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Step-by-step explanation:

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3 years ago
4.3t-2.1t-2.3=7.6<br> Solve equation
zlopas [31]
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5 0
4 years ago
There is no solution to the following system of equations.
sergij07 [2.7K]

Answer:

True

Step-by-step explanation:

We are given that system of equation

x<3

y>3

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x=3

y=3

When x=0

0<3

It is true equation therefore, the shaded region towards the origin.

When y=0

0>3

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From the graph

There is  no solution of system of equation.

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6 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
4 years ago
Read 2 more answers
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