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jok3333 [9.3K]
3 years ago
15

Plz help asap i will give u brainlist

Mathematics
1 answer:
nadya68 [22]3 years ago
4 0

Answer: (1,50) (2,100) (3,150) (4,200) (5,250) (6,300) (7,350) (8,400) (9,450)

Step-by-step explanation:

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Write an equation in standard form for the line that passes through the point (2,-3) and is perpendicular to the line y + 4 = -2
sineoko [7]

For this case we have that by definition, the standard form of a linear equation is given by:

ax + by = c

By definition, if two lines are perpendicular then the product of their slopes is -1. That is to say:

m_ {1} * m_ {2} = - 1

We have the following point-slope equation of a line:

y+4 = -\frac {2} {3}(x-12)

The slope is:

m_ {1} = - \frac {2} {3}

We find the slope m_ {2}of a perpendicular line:

m_ {2} = \frac {-1}{m_ {1}}\\m_ {2} = \frac {-1} {-\frac {2} {3}}\\m_ {2} = \frac{3} {2}

Thus, the equation is of the form:

y-y_ {0} = \frac {3} {2} (x-x_ {0})

We have the point through which the line passes:

(x_ {0}, y_ {0}) :( 2, -3)

Thus, the equation is:

y - (- 3) = \frac {3} {2} (x-2)\\y + 3 = \frac {3} {2} (x-2)

We manipulate algebraically:

y + 3 = \frac{3} {2} x- \frac {3} {2} (2)\\y + 3 = \frac{3} {2} x-3

We add 3 to both sides of the equation:

y + 3 + 3 = \frac {3} {2} x\\y + 6 = \frac {3} {2} x

We multiply by 2 on both sides of the equation:

2(y + 6) = 3x\\2y + 12 = 3x

We subtract 3x on both sides:

2y-3x + 12 = 0

We subtract 12 from both sides:

2y-3x = -12

ANswer:

-3x + 2y = -12

5 0
2 years ago
The angle measures 22°, 62°, 118°, and 158° are written on slips of paper. You choose two slips of paper at random.
mylen [45]
Answer: 1/3

There are C(4,2) = 6 ways to choose a pair of numbers.  Only 2 of those pairs are supplementary measures {22, 158} and {62, 118}. The probability is 2/6 = 1/3.


3 0
2 years ago
Explain how you would graph the following set of parametric equations by plotting points and describing the orientation.
lyudmila [28]
Concept:

First eliminate the t from x=3t and then put it in y=t² and then graph it. As, limit of t is not  restricted so t ∈ R(all real numbers)

As it is difficult to make graph here so I have solved it by hand and add all detail regarding it.

6 0
3 years ago
Read 2 more answers
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
PLZZ HELP ILL GIVE BRAINLIST
scZoUnD [109]

Answer:

V≈471.24

r Radius: 5

h Height: 6

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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