Order from least to greatest

-BARSIC- [3]

The answer is C because complimentary is equal to 90 so 24+66 equals 90 and 34+56 equal 90

3 0

3 years ago

given the points(-4,8) and (6,-12) determine the mid point of line segment connecting the points determine the distance sepratin

lbvjy [14]

Answer:

$midpoint = ( \frac{ - 4 + 6}{2} , \: \frac{ - 12 + 8}{2} ) \\ = (1, \: - 2) \\ \\ distance = \sqrt{ {( - 4 - 6)}^{2} + {(8 + 12)}^{2} } \\ = \sqrt{500} \\ = 22.4 \: units$

8 0

3 years ago

What is the equivalent expession of 24x-78y

pickupchik [31]

When you find the GCF (Greatest Common Factor), which is 6 you simplify both by 6 and get 4x-13y

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

3 years ago

Can someone please answer this question please I need it right now and please answer it correctly please show work answer number

schepotkina [342]

Answer:

24) A

22) B

Step-by-step explanation:

24)

Original Price: d

Sale: 30%

A sale means that a value is subtracted from the original price

d - 30% = 0.7d

All options given:

A) 0.3d

B) 0.7d

C) d - 0.3d = 0.7d

D) (1.0 - 0.3)d = 0.7(d) = 0.7d

All of these options are equal to the price after the sale, except A.

A is the answer to question 24

22) m > 0, n > 0 Use an example to represent m and n

m = 1, n = 2 = -0.5

-0.5 is a rational, negative number

The answer to question 22 is B

Hope this helps :)

5 0

3 years ago