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Phoenix [80]
3 years ago
14

A person in a tow truck traveled 7 miles south and then 6 miles east. The person realized she was lost,

Mathematics
1 answer:
Rufina [12.5K]3 years ago
7 0
In my understanding, the component form of a vector is the ordered pair that describes the changes in the x- and y-values.

Therefore to get it
you must add all the
x
(0 + (-6) + (-6) + 6)
= -6

Then add all y
(7 + 7 + (-11) + (-11)
= -8

Answer = (-6,-8)

You might be interested in
Iq scores are normally distributed with a mean of 100 and a standard deviation of 15 what is the probability that a randomly sel
a_sh-v [17]
So we are given the mean and the s.d.. The mean is 100 and the sd is 15 and we are trying the select a random person who has an I.Q. of over 126. So our first step is to use our z-score equation:

z = x - mean/s.d.

where x is our I.Q. we are looking for

So we plug in our numbers and we get:

126-100/15 = 1.73333

Next we look at our z-score table for our P-value and I got 0.9582

Since we are looking for a person who has an I.Q. higher than 126, we do 1 - P. So we get

1 - 0.9582 = 0.0418

Since they are asking for the probability, we multiply our P-value by 100, and we get 

0.0418 * 100 = 4.18%

And our answer is

4.18% that a randomly selected person has an I.Q. above 126
Hopes this helps!
6 0
3 years ago
Is this right
pychu [463]

Answer:

360 = x+123+ 90

x = 147

Step-by-step explanation:

The three angles form a circle which is 360 degrees

360 = x+123+ 90

Subtract 123 and 90 from each side

360 -123 - 90 = x

147 = x

6 0
3 years ago
Read 2 more answers
I am greater than 4 tens and less than 5 tens I have 9 ones
aliina [53]

4 tens = 40 and 5 tens = 50 … therefore the number has to be between 40 and 50

So: 40 < x <50 … this says that x is greater than 40 and less than 50

 

So the possible numbers would be: 41,42,43,44,45,46,47,48,49

Now the number must have 9 ones … which mean the only number would be 49

4 0
3 years ago
If angle DAB measures 34 degrees, what is the measure of arc DB
ollegr [7]
Arc DB will also measure 34 degrees.
4 0
3 years ago
Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
Schach [20]

Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

Solution :

a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

4 0
3 years ago
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