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Nana76 [90]
3 years ago
5

Help Box plots don’t make sense!!

Mathematics
1 answer:
Mashcka [7]3 years ago
7 0

Try counting the counter parts upward then count sideways

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HELP?! PLEASE!! SUPER DESPERATE!! How many gallons of a 60% antifreeze solution must be mixed with a 60 gallons of 15% antifreez
Ket [755]

bellamore

Middle SchoolMathematics 5+3 pts

x (.60) + 60 (.15) = (x+60) .50

distribute

.60x +9 = .5x +30

subtract  .5x from each side

.1x +9 = 30

subtract 9 from each side

.1x = 21

 divide by .1

x = 210

You will need 210 gallons of 60% antifreeze


8 0
3 years ago
PLEASE ANSWER REAL
Blababa [14]

Answer:

How come my answer gets deleted when a kid spammed -_-

6 0
3 years ago
The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q
Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

\frac{\sum f}{4}=\frac{100}{4}=25

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF)_{p} = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h

     =34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75

The value of first quartile is 34.75.

Q₃ is at the position:

\frac{3\sum f}{4}=\frac{3\times100}{4}=75

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF)_{p} = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h

     =44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83

The value of third quartile is 44.83.

Then the inter-quartile range is:

IQR = Q_{3}-Q_{1}

        =44.83-34.75\\=10.08

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}

                                 =\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0
4 years ago
Mrs. Johnson ordered 4 exotic plants for her yard for $62.95 each. She is getting them all delivered to her house on Wednesday.
Alborosie

Answer:

$311.75

Step-by-step explanation:

6 0
4 years ago
Read 2 more answers
Find the values of x and y that maximize the objective function P=3x+2y for the graph. What is the maximum value?
AURORKA [14]

Answer:

(9, 0)

Step-by-step explanation:

Maximum or minimum value occurs at the Corner. The points given are (0, 8), (5, 4) and (9, 0).

Substitute (0, 8) in the objective function.

We get P = 3(0) + 2(8) = 16

Now for (x , y) = (5, 4)

P = 3(5) + 2(4) = 15 + 8 = 23

At (9, 0) we get P = 3(9) + 2(0) = 27.

Clearly, we have the maximum value at (9, 0).

And the maximum value is 27.

3 0
3 years ago
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