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olga nikolaevna [1]
2 years ago
7

teresa stated that the heights of the students in her class were not a function of their ages. Which reasoning could justify Ter

esa’s statement? Two students are the same age but have different heights. Two students have the same height but different ages. No two students are the same age or height.
Mathematics
2 answers:
Masja [62]2 years ago
7 0

The answer is Two students are the same age but have different heights. Hope this helps!

PSYCHO15rus [73]2 years ago
6 0

A) Two students are the same age but have different heights.

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One of the factors of the polynomial x3 − 5x2 is x + 3. What is the other factor?
blondinia [14]

Answer:

C. x² − 8x + 24 − 72/(x+3)

Step-by-step explanation:

See attached picture for long division method.

Logically, we know x³ − 5x² factors to x² (x − 5).  Since x + 3 isn't a factor, we know the remainder isn't 0.  So we can narrow the options down to A or C.

One way to find the remainder is through long division.  Or, since this is multiple choice, we multiply the options by x + 3 and see which one results in an answer of x³ − 5x².

(x + 3) (x² − 8x + 24 − 72/(x+3))

(x + 3) (x² − 8x + 24) − 72

x³ − 8x² + 24x + 3x² − 24x + 72 − 72

x³ − 5x²

6 0
2 years ago
Sheryl creates a scatter plot to analyze how an increase in the outside temperature above 52°F affects the sale of hot chocolate
lora16 [44]

Answer:

1 and 2

Step-by-step explanation:

3 0
2 years ago
A semicircular flower bed has a diameter of 3 metres. What is the area of the flower bed?
Yuri [45]

The area of circle is

A_{\text{circle}}=\pi r^2.

Then the area of semicircular region is

A_{\text{semicircular region}}=\dfrac{1}{2} \pi r^2.

If r=3 m, you have

A_{\text{semicircular region}}=\dfrac{1}{2} \pi \cdot 3^2=\dfrac{9\pi}{2}=4.5\pi \approx 14.13 sq. m.

Answer: A_{\text{semicircular region}}=4.5\pi \approx 14.13 sq. m.

3 0
2 years ago
Read 2 more answers
In your sock drawer you have 5 blue, 7 gray, and 2 black socks. Half asleep one morning you grab 2 socks at random and put them
pentagon [3]

The question is incomplete! The complete question along with answers and explanation is provided below!

In your sock drawer you have 5 blue, 7 gray, and 2 black socks. Half asleep one morning you grab 2 socks at random and put them on. Find the probability you end up wearing the following socks. (Round your answers to four decimal places.)

a) 2 blue socks

b) no gray socks

c) at least 1 black sock

d) a green sock

e) matching socks

Answer:

a) P(2 blue socks) = 0.1099 = 10.99%

b) P(no gray socks) = 0.2307 = 23.07%

c) P(at least 1 Black sock) = 0.2748 = 27.48%

d) P(green sock) = 0%

e) P(Matching socks) = 0.3516 = 35.16%

Step-by-step explanation:

Given Information:

5 Blue socks

7 Gray socks

2 black socks

Total socks = 5 + 7 + 2 = 14

a) The probability of wearing 2 blue socks

P(2 blue socks) = P(B1 and B2)

P(B1) = no. of blue socks/total no. of socks

P(B1) = 5/14 = 0.3571

Now there are 4 blue socks remaining and total 13 socks remaining

P(B2|B1) = 4/13 = 0.3077

P(B1 and B2) = 0.3571*0.3077 = 0.1099 = 10.99%

b) The probability of wearing no gray socks

5 Blue socks + 2 black socks = 7 socks are not gray

P(no gray socks) = P(Not G1 and Not G2)

P(Not G1) = no. socks that are not grey/ total no. of socks

P(Not G1) = 7/14 = 0.5

Now there are 6 socks remaining that are not gray and total 13 socks remaining

P(Not G2 | Not G1) = 6/13 = 0.4615

P(Not G1 and Not G2) = 0.5*0.4615 = 0.2307 = 23.07%

c) The probability of wearing at least 1 black sock

5 Blue socks + 7 Gray socks = 12 socks are not black

P(at least 1 Black) = 1 - P(Not B1 and Not B2)

P(Not B1) = no. socks that are not black/ total no. of socks

P(Not B1) = 12/14 = 0.8571

Now there are 11 socks remaining that are not black and total 13 socks remaining

P(Not B2 | Not B1) = 11/13 = 0.8461

P( Not B1 and Not B2) = 0.8571*0.8461 = 0.7252

P(at least 1 Black) = 1 - P( Not B1 and Not B2)

P(at least 1 Black) = 1 - 0.7252 = 0.2748 = 27.48%

d) The probability of wearing a green sock

There are 0 green socks, therefore

P(Green) = 0/14 = 0%

e) The probability of wearing matching socks

P(Matching socks) = P(2 Blue socks) + P(2 Gray socks) + P(2 Black socks)

P(2 Blue socks) already calculated in part a

P(2 Blue socks) = P(B1 and B2) = 0.1099

For Gray socks

P(G1) = no. of gray socks/ total no. of socks

P(G1) = 7/14 = 0.5

Now there are 6 gray socks remaining and total 13 socks remaining

P(G2 | G1) = 6/13 = 0.4615

P(2 Gray socks) = P(G1 and G2) = 0.5*0.4615 = 0.2307

For Black socks

P(B1) = no. of black socks/ total no. of socks

P(B1) = 2/14 = 0.1428

Now there is 1 black sock remaining and total 13 socks remaining

P(B2 | B1) = 1/13 = 0.0769

P(2 Black socks) = P(B1 and B2) = 0.1428*0.0769 = 0.0110

P(Matching socks) = P(2 Blue socks) + P(2 Gray socks) + P(2 Black socks)

P(Matching socks) = 0.1099 + 0.2307 + 0.0110 = 0.3516 = 35.16%

7 0
2 years ago
Use the table to answer the question. House A $124,270 Annual appreciation 4% House B $114,270 Annual appreciation 5% In which o
Varvara68 [4.7K]
For the house A we have:
 f (x) = 124270 (1.04) ^ x
 Evaluating for 7, 8, 9 and 10 we have:
 f (7) = 124270 (1.04) ^ 7 = 163530.8422
 f (8) = 124270 (1.04) ^ 8 = 170072.0759
 f (9) = 124270 (1.04) ^ 9 = 176874.9589
 f (10) = 124270 (1.04) ^ 10 = 183949.9573

 For house B we have:
 
f (x) = 114270 (1.05) ^ x
 Evaluating for 7, 8, 9 and 10 we have:
 f (7) = 114270 (1.05) ^ 7 = 160789.3653
 f (8) = 114270 (1.05) ^ 8 = 168828.8336
 f (9) = 114270 (1.05) ^ 9 = 177270.2752
 f (10) = 114270 (1.05) ^ 10 = 186133.789

 We observe that for years 7 and 8 the value of house A is greater than the value of house B.

 Answer:
 
7 and 8
4 0
2 years ago
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