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Assoli18 [71]
3 years ago
11

Help with 15 thanks

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
8 0
I think the answer would be C because variability means average.
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A line passes through the point (–7, 5) and has a slope of 1/5. Which is another point that the line passes through?
ICE Princess25 [194]
<span>A line passes through the point (–7, 5) and has a slope of 1/5.
So this line would be y=1/5x+b
If we substitute x and y into this </span>formula
We can get b=5+7/5=32/5
Then y=1/5x+32/5
If x=0, then y=32/5
So another point on this line is (0,32/5)
8 0
3 years ago
Read 2 more answers
The rental car agency has 30 cars on the lot. 10 are in great shape, 16 are in good shape, and 4 are in poor shape. Four cars ar
Korolek [52]

Complete Question:

The rental car agency has 30 cars on the lot. 10 are in great shape, 16 are in good shape, and 4 are in poor shape. Four cars are selected at random to be inspected. Do not simplify your answers. Leave in combinatorics form. What is the probability that:

a. Every car selected is in poor shape

b. At least two cars selected are in good shape.

c. Exactly three cars selected are in great shape.

d. Two cars selected are in great shape and two are in good shape.

e. One car selected is in good shape but the other 3 selected are in poor shape.

Answer:

a

   P_A = \frac{^4 C_4}{^{30}C_{4}}

b

  P_B = \frac{[^{16} C_2 *^{14} C_2 ] +[^{16} C_3 *^{14} C_1 ] + ^{16} C_4}{^{30}C_{4}}

c

  P_C = \frac{^{10} C_3 *^{20} C_1 }{^{30}C_{4}}

d

   P_D = \frac{^{10} C_2 *^{16} C_2 }{^{30}C_{4}}

e

   P_E = \frac{^{16} C_1 *^{4} C_3 }{^{30}C_{4}}

Step-by-step explanation:

From the question we are told that

 The number of car in the parking lot is  n =  30  

  The number of cars in great shape is  k =  10

   The number of cars in good shape is r = 16

    The number of cars in poor shape is q = 4

The number of cars that were selected at random is N= 4

Considering question a

    Generally the number of way of selecting four cars that are in a poor shape from number of cars that are in poor shape is

            ^{q} C_{N}

=>        ^{4} C_{4}

Here C stands for combination.

 Generally the number of way of selecting four cars that are in a poor shape from total number of cars  in the parking lot is

          ^{n} C_{N}

=>      ^{30} C_{4}

Generally the probability that every car selected is in poor shape  is mathematically represented as

       P_A = \frac{^4 C_4}{^{30}C_{4}}

Considering question b

Generally the number of way of selecting 2 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{2}

=>        ^{16} C_{2}

Here C stands for combination.

 Generally the number of way of selecting the remaining 2 cars  from the remaining number of cars  in the parking lot is

          ^{n-r} C_{2}

=>      ^{30-16} C_{2}

=>      ^{14} C_{2}

Generally the number of way of selecting 3 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{3}

=>        ^{16} C_{3}

 Generally the number of way of selecting the remaining 1 cars  from the remaining number of cars  in the parking lot is

          ^{n-r} C_{1}

=>      ^{30-16} C_{1}

=>      ^{14} C_{1}

Generally the number of way of selecting 4 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{4}

=>        ^{16} C_{4}

Generally the probability that at least two cars selected are in good shape

       P_B = \frac{[^{16} C_2 *^{14} C_2 ] +[^{16} C_3 *^{14} C_1 ] + ^{16} C_4}{^{30}C_{4}}

Considering question c

Generally the number of way of selecting 3 cars that are in great shape from number of cars that are in great shape is      

            ^{k} C_{3}

=>        ^{10} C_{3}

 Generally the number of way of selecting the remaining 1 cars  from the remaining number of cars  in the parking lot is

          ^{n-k} C_{1}

=>      ^{30-10} C_{1}

=>      ^{20} C_{1}

Generally the probability of selecting exactly three cars selected are in great shape is

        P_C = \frac{^{10} C_3 *^{20} C_1 }{^{30}C_{4}}

Considering question d

Generally the number of way of selecting 2 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{2}

=>        ^{16} C_{2}

Generally the number of way of selecting 2 cars that are in great shape from number of cars that are in great shape is      

            ^{k} C_{2}

=>        ^{10} C_{2}

Generally the probability that two cars selected are in great shape and two are in good shape.

              P_D = \frac{^{10} C_2 *^{16} C_2 }{^{30}C_{4}}

Considering question e

Generally the number of way of selecting 1 cars that is in good shape from number of cars that are in good shape is

            ^{r} C_{1}

=>        ^{16} C_{1}

    Generally the number of way of selecting 3 cars that are in a poor shape from number of cars that are in poor shape is

            ^{q} C_{3}

=>        ^{4} C_{3}

Generally the probability that one car selected is in good shape but the other 3 selected are in poor shape is

         P_E = \frac{^{16} C_1 *^{4} C_3 }{^{30}C_{4}}

4 0
3 years ago
Zara can travel 25km in one and a half hours on her bicycle. how long will it take her to travell 15km?
satela [25.4K]
8.3 hours it would take cann to travele
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30 degrees north of west is what heading?
Zinaida [17]
I believe your answer is North West.
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Set up a right triangle model for this problem and solve by using the reference table trigonometric ratio that applies. Follow t
RSB [31]
In reference to the angle of elevation (41 degrees), the adjacent side is 60 and the opposite side is the unknown.
Using tangent ratio:
tan 41 = opp/60
60 tan41 = opp
opp side = 52.2yards
LETTER B
5 0
3 years ago
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