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kykrilka [37]
3 years ago
11

What is the location of the ordered pair (-3,2) on the coordinate plane

Mathematics
2 answers:
inn [45]3 years ago
7 0
You have to go 3 squares to the left (staying in the x coordinate of 0) and after that you go 2 squares up. Hope this helps. :)
Ber [7]3 years ago
5 0
Let me draw it for you.

It would look like this.


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What is the value of 11−3m , when m = 2? (Please do it step by step because I dont know how to do this ;[ )
STatiana [176]
Substitute the variable for the value given.

11 - 3m becomes 11 - 3(2)

Now do the order of operations.

11 - 3(2) =
11 - 6 =
5

So, the answer is 5.
6 0
2 years ago
In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
Marrrta [24]

Answer:

a) Bi [P ( X >=15 ) ] ≈ 0.9944

b) Bi [P ( X >=30 ) ] ≈ 0.3182

c)  Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) Bi [P ( X >40 ) ] ≈ 0.0046  

Step-by-step explanation:

Given:

- Total sample size n = 745

- The probability of success p = 0.037

- The probability of failure q = 0.963

Find:

a. 15 or more will live beyond their 90th birthday

b. 30 or more will live beyond their 90th birthday

c. between 25 and 35 will live beyond their 90th birthday

d. more than 40 will live beyond their 90th birthday

Solution:

- The condition for normal approximation to binomial distribution:                                                

                    n*p = 745*0.037 = 27.565 > 5

                    n*q = 745*0.963 = 717.435 > 5

                    Normal Approximation is valid.

a) P ( X >= 15 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]

 - Then the parameters u mean and σ standard deviation for normal distribution are:

                u = n*p = 27.565

                σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522

- The random variable has approximated normal distribution as follows:

                X~N ( 27.565 , 5.1522^2 )

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= -2.5358 ) = 0.9944

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944

Hence,

                Bi [P ( X >=15 ) ] ≈ 0.9944

b) P ( X >= 30 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= 0.37556 ) = 0.3182

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182

Hence,

                Bi [P ( X >=30 ) ] ≈ 0.3182  

c) P ( 25=< X =< 35 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )

                N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

- Now use the Z-score table to evaluate the probability:

                P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

               N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

Hence,

                Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) P ( X > 40 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )

                N [ P ( X > 41 ) ] = P ( Z > 2.60762 )

- Now use the Z-score table to evaluate the probability:

               P ( Z > 2.60762 ) = 0.0046

               N [ P ( X > 41 ) ] =  P ( Z > 2.60762 ) = 0.0046

Hence,

                Bi [P ( X >40 ) ] ≈ 0.0046  

4 0
3 years ago
Question No. 4 The area of an equilateral triangle is half the area of a rhombus which in turn is equal to one third the area of
____ [38]

The ratio of the length of each side of the triangle to that of the hexagon is

<h3>What is equilateral triangle?</h3>

The triangle which has all the three sides equal in dimension is called the equilateral triangle.

According to the question,

Area of triangle = Area of rhombus = Area of hexagon

1/2 base x height = 1/2 diagonal1 xdiagonal2 = (3√3 / 2) side²

The triangle has equal base and height = a.

The hexagon has each side measuring b.

1/2 a² = (3√3 / 2) b²

Ratio of sides of triangle and hexagon is

a²/b² = 3√3

a/b = √ 5.19615

a/b = 2.2795

Thus, the ratio of each side of the triangle to that of the hexagon is 2.2795.

Learn more about equilateral triangle.

brainly.com/question/3461022

#SPJ1

4 0
1 year ago
Someone please help me ASAP!!!! I really need it
MatroZZZ [7]
^ this guy is a scammer don’t click the link
7 0
3 years ago
On three days, Ben throws darts at a target.
Genrish500 [490]

Answer:

0.6 ;

0.7

Reaukt in B will give a better estimate because it has a larger sample size (120) than the result on A ,with sample size (30)

Step-by-step explanation:

Probability of hitting the target using Tuesday's result :

P(hitting the target):

Probability = required outcome / Total possible outcomes

Required outcome = number of hits = 18

Total possible outcomes = number of throws =. 30

Hence, Probability of hitting the target using Tuesday's result : 18 / 30 = 0.6

B.)

Probability of hitting the target using Total result:

Required outcome = number of hits = 84

Total possible outcomes = number of throws =. 120

= 84 / 120

= 0.7

4 0
2 years ago
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