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Lunna [17]
3 years ago
15

Javier's fuel tank holds 12 3/4 gallons of gasoline when completely full. He had some gas in the tank and added 10.3 gallons of

gasoline to fill it completely.
Mathematics
1 answer:
GenaCL600 [577]3 years ago
6 0

Answer:

2.45 gallons of gasoline were in the tank before.

Step-by-step explanation:

Consider the provided information.

The tank can hold 12\frac{3}{4} gallons of gasoline when completely full. He had some gas in the tank and added 10.3 gallons of gasoline to fill it completely.

We need to find: How many gallons of gasoline were in the tank before Javier added some?

For this, we need to subtract added gallons of gasoline from the capacity of the tank.

12\frac{3}{4}-10.3=\frac{51}{4}-10.3

=12.75-10.3

=2.45

Hence, 2.45 gallons of gasoline were in the tank before.

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First, we have to get the fractions to common denominators.

4/5 = 16/20
1/4 = 5/20

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She put 1 9/20 extra cups in.
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I have many students are represented in the data set?
nalin [4]

Answer:

a) 19 students

b) the mean 3 9/19

the median 3

the mode 3

c) the range 6

Step-by-step explanation:

The data set shows that

0 points gets 1 student,

1 point gets 1 student,

2 points get 2 students,

3 points get 6 students,

4 points get 4 students,

5 points get 3 students and

6 points get 2 students.

a) There are 1+1+2+6+4+3+2=19 students.

b) The mean is

\dfrac{1\cdot 0+1\cdot 1+2\cdot 2+6\cdot 3+4\cdot 4+3\cdot 5+2\cdot 6}{19}=\dfrac{66}{19}=3\dfrac{9}{19}.

The average score is 3 9/19 points.

The median is 10th term in the data set - 3 points (means the middle score in the data set)

The mode is 3 points (means most happened score)

c) The range of the data is 6-0=6 points.

7 0
3 years ago
Sean b= 5,79; c= 10,4,el angulo A= 54,46°, el angulo C mide ?
Rzqust [24]

Answer:

C \approx 91.732^{\circ}

Step-by-step explanation:

(This exercise is presented in Spanish and for that reason explanation will be held in such language)

El lado restante se determina por la Ley del Coseno:

a = \sqrt{b^{2}+c^{2}-2\cdot b\cdot c \cdot \cos A}

a = \sqrt{5.79^{2}+10.4^{2}-2\cdot (5.79)\cdot (10.4)\cdot \cos 54.46^{\circ}}

a \approx 8.466

Finalmente, el angulo C se halla por medio de la misma ley:

\cos C = - \frac{c^{2}-a^{2}-b^{2}}{2\cdot a \cdot b}

\cos C = -\frac{10.4^{2}-8.466^{2}-5.79^{2}}{2\cdot (8.466)\cdot (5.79)}

\cos C = -0.030

C \approx 91.732^{\circ}

3 0
3 years ago
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