Please help me figure this out explain it to me

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

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