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Elenna [48]
3 years ago
6

8th grade students held a fundraiser dinner. they paid a flat rate of $125 for the use of a building plus $13 for each student w

ho attended. the total cost was $944. how many students attended?
Mathematics
2 answers:
LekaFEV [45]3 years ago
8 0

Answer: 63 students

Step-by-step explanation: Subtract $944-$125 = $819. After we get this answer we divide it by $13. $819 / $13= 63. Therefore, 63 students attended the dinner.

daser333 [38]3 years ago
7 0

Answer: 63

Step-by-step explanation:

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SCORPION-xisa [38]

Answer:

Hi i think that the correct answer is B

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Muriel and Ramon bought school
Dafna11 [192]

Answer:

the box of pencils are $2.50

the erasers are $1.50

Step-by-step explanation:

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7 0
3 years ago
With a little care, you can find the median and the quartiles from the histogram. what are these numbers? you can also find the
coldgirl [10]

Complete Question

The histogram from this question is shown on the first uploaded image

Answer:

The first quartile is  1st Q = 19^{th} girl (subject ) = 1 \serving\ of\ fruit\ per\ day

The median is   Median  = 38 ^{th} \ girl (subject) =  2 \serving\ of\ fruit\ per\ day

The third quartile  is 3rd Q =  56^{th} \  girl (subject) = 4 \serving\ of\ fruit\ per\ day  

The  mean is  \= x = 2.62

Step-by-step explanation:

From the question we are told that

   The total  number of girls is n =  74

Generally the median is mathematically represented as

       Median  =  \frac{\frac{n}{2}  +[ \frac{n}{2} + 1]  }{2}

So

          Median  =  \frac{\frac{74}{2}  +[ \frac{74}{2} + 1]  }{2}

=>         Median  =  \frac{37  +38 }{2}

=>         Median  =37.5 \  girl    

From the histogram 37.5 \approx 38^{th} \ girl fall under 2 fruits per day

Generally the first quartile is mathematically represented as

      1st \ Q =  \frac{\frac{n}{4} + [\frac{n}{4} + 1]  }{2}

=>    1st \ Q =  \frac{\frac{74}{4} + [\frac{74}{4} + 1]  }{2}

=>     1st \ Q = 19 \  girl

From the histogram 19^{th}girl fall under 1 fruit per day

 Generally the third quartile is mathematically represented as

     3rd\  Q =  \frac{\frac{n}{2}  + n }{2}

=>    3rd\  Q =  \frac{\frac{74}{2}  + 74 }{2}

=>    3rd\  Q =  55.5

From the histogram 55.5 \approx 56^{th} \ girl fall under 4 fruits per day

Generally from the histogram table the frequency  [number of subjects (girls) ]  and the servings of fruit per day can be represented as

                                                                                                               Total

x(servings per day )        0     1     2     3     4     5    6     7      8    

f (number of subjects )   15    11   15     11    8     5    3     3      3    \sum f  =  74

xf                                       0    11    30   33   32  25   18   21     24    \sum xf =  194

Generally the mean is mathematically represented as

      \= x = \frac{1}{ \sum f}  * [\sum xf]

=>     \= x = \frac{1}{ 74}  *194

=>      \= x = 2.62

4 0
3 years ago
Given f (X) = 4x +6 find f (x) = 54
Nikitich [7]

Answer:

x = 12

Step-by-step explanation:

Step 1: Define

f(x) = 4x + 6

f(x) = 54

Step 2: Substitute variables

54 = 4x + 6

Step 3: Solve for <em>x</em>

<u>Subtract 6 on both sides:</u> 48 = 4x

<u>Divide both sides by 4:</u> 12 = x

Step 4: Check

<em>Plug in x to verify it is a solution.</em>

f(12) = 4(12) + 6

f(12) = 48 + 6

f(12) = 54

3 0
3 years ago
Read 2 more answers
You are testing the claim that the mean GPA of night students is different from the mea GPA of day students. You sample 20 night
svetlana [45]

Answer:

As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

Step-by-step explanation:

Here n= 20

Sample mean GPA = x`= 2.84

Standard mean GPA = u= 2.55

Standard deviation = s=  0.45.

Level of Significance.= ∝ = 0.01

The hypothesis are formulated as

H0: u1=u2   i.e the GPA of night students is same as the mean GPA of day students

against the claim

Ha: u1≠u2

i.e the GPA of night students is different from the mea GPA of day students

For two tailed test  the critical value is  z ≥ z∝/2= ± 2.58

The test statistic

Z= x`-u/s/√n

z= 2.84-2.55/0.45/√20

z= 0.1441

As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

8 0
3 years ago
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