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4 years ago

Please help on 4-7 please I will give thanks or brainiest

Mathematics

2 answers:

Vladimir79 [104]4 years ago

7 0

4) 9:35 5) 3hours 40 minutes 6) 2:05 7) 11:24

kifflom [539]4 years ago

7 0

Hi! Rather than give you the answer to all four im going to walk you through the two diffrent types of questions you have - that way if this comes up again you can solve it on your own and have a sense of accomplishment - let me know if this helps you!!#47:15-2h= 5:15 now we subract the 20. Since 20 is bigger than 15 subtract 15. now we're at 5:00; minus 5 minutes. Well; what time at 4:00 is only 5 minutes to 5? 4:55. END TIME= 4:55
#6 Okay so this one is backwards from what we just did.
Its 7:55, and 5hrs and 50minutes have passed to get us there. 7-5=2. SO now we have 2:55. 55-50=5. START TIME= 2:05

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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

Please help me with this it would be very much appreciated

lara [203]

Answer:

423.5

Step-by-step explanation:

12.1 times 10 times 7 = 847

847/2 = 423.5

6 0

3 years ago

Open the parallel lines translation application. The right set of parallel lines is a translation right 10 units of the left set

Luden [163]

When a translation is included on the axes for any shape or line, lines that are otherwise identical cannot be compared to one another. in the same manner as the lines that I drew in parallel. These two lines do not compete with one another in any way. This is further explained below.

<h3>What is translation?</h3>

Generally, Changing words from one language to another is just one of the processes of translation. In other words, translation facilitates communication between different cultural groups. You may get a firsthand look at cultural phenomena that, seen through the perspective of your own culture, would seem alien and distant.

In conclusion, Lines that are otherwise similar to one another are unable to be compared to one another when an additional translation is put on the axes for any form or line. in exactly the same way as the lines that I drew parallel to one another. These two different lines are in no way intended to compete with one another.