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riadik2000 [5.3K]
2 years ago
8

Please help me figure this out explain it to me

Mathematics
1 answer:
jeka942 years ago
8 0

Answer:

40yd

Step-by-step explanation:

area of total minus unshaded portion

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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
2 years ago
A robot that makes _/6 of a boat per day will make 5 boats in 6 days
MatroZZZ [7]
For what I understand, if a robot was to make 5/6 of a boat per day, in 6 days it will have made 5 boats. The answer is 5
4 0
2 years ago
What is up !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
timofeeve [1]

Answer: the sky and planets

Step-by-step explanation:

the are above us

7 0
2 years ago
A(n) = -5 + 6(n - 1)a(n)=−5+6(n−1)a, left parenthesis, n, right parenthesis, equals, minus, 5, plus, 6, left parenthesis, n, min
DENIUS [597]

Answer:

The 12th term is 61

Step-by-step explanation:

I will assume that your a(n) = -5 + 6(n - 1) is correct; the rest is redundant (duplicative, unneeded).

To find the 12th term, substitute 12 for n in the above formula:

a(12) = -5 + 6(12 - 1) = -5 + 6(11) = 66 - 5, or 61

The 12th term is 61

6 0
2 years ago
A contractor is required by a county planning department to submit one, two, three, four, five, or six forms (depending on the n
Ratling [72]

Answer:

a)

k = \dfrac{1}{21}

b) 0.476

c) 0.667    

Step-by-step explanation:

We are given the following in the question:

Y = the number of forms required of the next applicant.

Y: 1, 2, 3, 4, 5, 6

The probability is given by:

P(y) = ky

a) Property of discrete probability distribution:

\displaystyle\sum P(y_i) = 1\\\\\Rightarrow k(1+2+3+4+5+6) = 1\\\\\Rightarrow k(21) = 1\\\\\Rightarrow k = \dfrac{1}{21}

b) at most four forms are required

P(y \leq 4) = \displaystyle\sum^{y=4}_{y=1}P(y_i)\\\\P(y \leq 4) = \dfrac{1}{21}(1+2+3+4) = \dfrac{10}{21} = 0.476

c) probability that between two and five forms (inclusive) are required

P(2\leq y \leq 5) = \displaystyle\sum^{y=5}_{y=2}P(y_i)\\\\P(2\leq y \leq 5) = \dfrac{1}{21}(2+3+4+5) = \dfrac{14}{21} = 0.667

8 0
3 years ago
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