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2 years ago

Which one is correct?

Mathematics

2 answers:

JulijaS [17]2 years ago

6 0

Answer:

85 (D.)

Step-by-step explanation:

The sum of all the degrees in a quadrilateral is 360, and since a triangle is half of a quadrilateral, the sum of the degrees in a triangle is 180.

With the given information, we can create the equation:

40+55+x=180

Simplifying on the left side gives us:

95+x=180

Subtracting 95 from both sides gives us:

x=85

<u>Therefore, the solution to your problem is 85.</u>

Hope this helped!

ddd [48]2 years ago

3 0

Answer:

85°

Step-by-step explanation:

Just add the known angles and subtract them from 180 to get the missing angle.

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The box plots show the weights, in pounds, of the dogs in two different animal shelters. Weights of Dogs in Shelter A 2 box plot

katovenus [111]

Answer:

1) The median weight for shelter A is greater than that for shelter B.

True, the median weight for shelter A is 21 pounds as compared to the 18 pounds median weight for shelter B.

4) The data for shelter B are a symmetric data set.

True, the median is exactly 2 pounds away from each quartile which means it is symmetric.

5) The interquartile range of shelter A is greater than the interquartile range of shelter B.

True, the interquartile range of shelter A is 11 pounds as compared to the 4 pounds interquartile range of shelter B.

Step-by-step explanation:

We are given two box-plots, a box plot basically shows 5 statistical characteristics of a data set and they are

1. minimum value

2. Lower quartile

3. Median value

4. Upper quartile

5. Maximum value

Box-plot of dogs in shelter A:

The whiskers range from 8 to 30

Which means that the range is = 30 - 8 = 22

The box ranges from 17 to 28

Which means that the interquartile range is = 28 - 17 = 11

21 - 17 = 4

28 - 21 = 9

The median is not equal distance away from each quartile which means that the box-plot is not symmetric.

A line divides the box at 21

Which means that the median is = 21

Box-plot of dogs in shelter B:

The whiskers range from 10 to 28

Which means that the range is = 28 - 10 = 20

The box ranges from 16 to 20

Which means that the interquartile range is = 20 - 16 = 4

18 - 16 = 2

20 - 18 = 2

The median is exactly 2 pounds away from each quartile which means it is symmetric.

A line divides the box at 18

Which means that the median is = 18

Which is true of the data in the box plots?

1) The median weight for shelter A is greater than that for shelter B.

True, the median weight for shelter A is 21 pounds as compared to the 18 pounds median weight for shelter B.

2) The median weight for shelter B is greater than that for shelter A.

False

3) The data for shelter A are a symmetric data set.

False

4) The data for shelter B are a symmetric data set.

True

5) The interquartile range of shelter A is greater than the interquartile range of shelter B.

True, the interquartile range of shelter A is 11 pounds as compared to the 4 pounds interquartile range of shelter B.

5 0

3 years ago

Help please this is a quiz

aleksley [76]

Answer:

x^3 - 3x^2 +2x

Step-by-step explanation:

Use the distributive property

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2 years ago

Read 2 more answers

F(x) = 2x^3 + 8

riadik2000 [5.3K]

Answer:32-10x

Step-by-step explanation:

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3 years ago

In the figure, the ratio of the perimeter of rectangle ABDE to the perimeter of triangle BCD is 2/3 1 3/2 2 . The area of polygo

adelina 88 [10]

It would b 2/3 and the other one 2/4

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3 years ago

b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum

Mariana [72]

Answer:

${52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in ${52 \choose 7}$ ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in ${7 \choose 2}$ ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in ${45 \choose 7}$ . Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in ${7 \choose 2}$ ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in ${38 \choose 7}$ . And again, we have to choose which ones are the hidden ones, which can be chosen in ${7 \choose 2}$ ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in ${31 \choose 7}$ ways. And choosing which ones are the hidden ones for this player can be done in ${7 \choose 2}$ ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

${52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

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3 years ago