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Inessa [10]
3 years ago
9

Point A (-8, 7) and Point B (6,-9) form a line segment on a coordinate plane. What is the midpoint of line segment AB

Mathematics
2 answers:
BartSMP [9]3 years ago
6 0

Answer:

-1,-1

Step-by-step explanation:

it is the midpint (-1,-1)

Allisa [31]3 years ago
4 0

Answer:

( -1 , -1 )

Step-by-step explanation:

( -8 + 6 / 2 ) , ( 7 + -9 / 2)

( -2 / 2) , ( -2 / 2)

( -1 , -1 )

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95×2 the answer to problem is​
KengaRu [80]

95 times 2 is 190.

90 times 2 is 180, and 5 times 2 is 10.

180 + 10 = 190.

3 0
3 years ago
Read 2 more answers
A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp
abruzzese [7]

Step-by-step explanation:

<em>"A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp."</em>

<em />

<em>"(a) During a morning activity of the camp, these 12 players have to randomly group into six pairs of two players each."</em>

<em>"(i) Find the total number of possible ways that these six pairs can be formed."</em>

The order doesn't matter (AB is the same as BA), so use combinations.

For the first pair, there are ₁₂C₂ ways to choose 2 people from 12.

For the second pair, there are ₁₀C₂ ways to choose 2 people from 10.

So on and so forth.  The total number of combinations is:

₁₂C₂ × ₁₀C₂ × ₈C₂ × ₆C₂ × ₄C₂ × ₂C₂

= 66 × 45 × 28 × 15 × 6 × 1

= 7,484,400

<em>"(ii) Find the probability that each pair contains players of the same gender only. Correct your final answer to 4 decimal places."</em>

We need to find the number of ways that 6 boys can be grouped into 3 pairs.  Using the same logic as before:

₆C₂ × ₄C₂ × ₂C₂

= 15 × 6 × 1

= 90

There are 90 ways that 6 boys can be grouped into 3 pairs, which means there's also 90 ways that 6 girls can be grouped into 3 pairs.  So the probability is:

90 × 90 / 7,484,400

= 1 / 924

≈ 0.0011

<em>"(b) During an afternoon activity of the camp, 6 players are randomly selected and 6 one-on-one matches with the coach are to be scheduled.</em>

<em>(i) How many different schedules are possible?"</em>

There are ₁₂C₆ ways that 6 players can be selected from 12.  From there, each possible schedule has a different order of players, so we need to use permutations.

There are 6 options for the first match.  After that, there are 5 options for the second match.  Then 4 options for the third match.  So on and so forth.  So the number of permutations is 6!.

The total number of possible schedules is:

₁₂C₆ × 6!

= 924 × 720

= 665,280

<em>"(ii) Find the probability that the number of selected male players is higher than that of female players given that at most 4 females were selected. Correct your final answer to 4 decimal places."</em>

If at most 4 girls are selected, that means there's either 0, 1, 2, 3, or 4 girls.

If 0 girls are selected, the number of combinations is:

₆C₆ × ₆C₀ = 1 × 1 = 1

If 1 girl is selected, the number of combinations is:

₆C₅ × ₆C₁ = 6 × 6 = 36

If 2 girls are selected, the number of combinations is:

₆C₄ × ₆C₂ = 15 × 15 = 225

If 3 girls are selected, the number of combinations is:

₆C₃ × ₆C₃ = 20 × 20 = 400

If 4 girls are selected, the number of combinations is:

₆C₂ × ₆C₄ = 15 × 15 = 225

The probability that there are more boys than girls is:

(1 + 36 + 225) / (1 + 36 + 225 + 400 + 225)

= 262 / 887

≈ 0.2954

7 0
2 years ago
Pleeeeeseee have to finish before 3:00
aleksley [76]

Answer:

175.9

Step-by-step explanation:

AL=2πrh=2·π·4·7≈175.92919 rounded to the nearest tenth would be 175.9

5 0
2 years ago
I need a answer asap
Murrr4er [49]

Answer:

As the weight increases, the price increases.

6 0
3 years ago
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The sign of the product of –35 and –625 is
Anarel [89]

Step-by-step explanation:

The mathematical sign for the operation is subtraction.

<em>To</em><em> </em><em>Solve</em><em> </em><em>This</em><em> </em><em>You</em><em> </em><em>Will</em><em> </em><em>Need</em><em> </em><em>To</em><em>;</em>

<em>1</em><em>]</em><em> </em><em>The</em><em> </em><em>sign</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>product</em><em> </em><em>of</em><em> </em><em>-</em><em>3</em><em>5</em><em> </em><em>and </em><em>-</em><em>6</em><em>2</em><em>5</em><em> </em><em>is</em><em>;</em>

<em>It's</em><em> </em><em>going </em><em>to</em><em> </em><em>be</em><em>;</em>

<em>-</em><em>3</em><em>5</em><em>-</em><em>-</em><em>6</em><em>2</em><em>5</em>

<em>The</em><em> </em><em>negative</em><em> </em><em>sign</em><em> </em><em>will</em><em> </em><em>turn</em><em> </em><em>into </em><em>addition </em><em>sign</em><em> </em><em>before </em><em>you </em><em>can </em><em>operate</em><em> </em><em>it</em><em>.</em><em> </em><em>So </em><em>it</em><em>'s</em><em> </em><em>going</em><em> </em><em>to </em><em>be</em><em>;</em>

<em>-</em><em>3</em><em>5</em><em>+</em><em>6</em><em>2</em><em>5</em><em>=</em><em>5</em><em>9</em><em>0</em><em>(</em><em>positive</em><em>)</em>

<em>It's</em><em> </em><em>just</em><em> </em><em>like</em><em> </em><em>subtracting </em><em>3</em><em>5</em><em> </em><em>from</em><em> </em><em>6</em><em>2</em><em>5</em><em>.</em>

<em>2</em><em>]</em><em> </em><em>The </em><em>sign</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>product</em><em> </em><em>of </em><em>2</em><em>6</em><em>3</em><em> </em><em>and</em><em> </em><em>0</em><em> </em><em>is</em><em>;</em>

<em>2</em><em>6</em><em>3</em><em>-</em><em>0</em>

<em>With </em><em>this</em><em> </em><em>no</em><em> </em><em>qua</em><em>n</em><em>t</em><em>i</em><em>t</em><em>y</em><em> </em><em>is</em><em> </em><em>going</em><em> </em><em>to</em><em> </em><em>be</em><em> </em><em>subtracted </em><em>so</em><em> </em><em>the</em><em> </em><em>2</em><em>6</em><em>3</em><em> </em><em>will</em><em> </em><em>remain</em><em> </em><em>the</em><em> </em><em>same</em><em>.</em><em> </em>

<em>So</em><em> </em><em>2</em><em>6</em><em>3</em><em>-</em><em>0</em><em>=</em><em>2</em><em>6</em><em>3</em><em>(</em><em>positive</em><em>)</em>

<em>3</em><em>]</em><em> </em><em>The</em><em> </em><em>sign</em><em> </em><em>of </em><em>the</em><em> </em><em>product</em><em> </em><em>of</em><em> </em><em>-</em><em>2</em><em>1</em><em> </em><em>and</em><em> </em><em>4</em><em>5</em><em>1</em><em> </em><em>is</em><em>;</em>

<em>-</em><em>2</em><em>1</em><em>-</em><em>4</em><em>5</em><em>1</em>

<em>With </em><em>this</em><em> </em><em>you </em><em>won't</em><em> </em><em>subtract</em><em> </em><em>but</em><em> </em><em>rather</em><em> </em><em>you'll</em><em> </em><em>add</em><em> </em><em>and</em><em> </em><em>when</em><em> </em><em>you</em><em> </em><em>add</em><em> </em><em>the</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>going </em><em>to</em><em> </em><em>be</em><em> </em><em>-</em><em>4</em><em>7</em><em>2</em>

<em>-</em><em>4</em><em>7</em><em>2</em><em>(</em><em>negative</em><em>)</em>

<em>4</em><em>]</em><em> </em><em>The</em><em> </em><em>sign </em><em>of</em><em> </em><em>the</em><em> </em><em>product </em><em>of</em><em> </em><em>-</em><em>3</em><em>5</em><em>0</em><em> </em><em>and</em><em> </em><em>8</em><em>9</em><em> </em><em>is</em><em>;</em>

<em>-</em><em>3</em><em>5</em><em>0</em><em>-</em><em>8</em><em>9</em>

<em>With </em><em>this</em><em> </em><em>you</em><em> </em><em>will</em><em> </em><em>add</em><em> </em><em>it</em><em> </em><em>but</em><em> </em><em>you </em><em>won't</em><em> </em><em>subtract</em><em> </em><em>the</em><em> </em><em>negatives </em><em>so</em><em> </em><em>it's</em><em> </em><em>going</em><em> </em><em>to </em><em>be</em><em>;</em>

<em>-</em><em>3</em><em>5</em><em>0</em><em>-</em><em>8</em><em>9</em><em>=</em><em>-</em><em>4</em><em>3</em><em>9</em><em>(</em><em>negative</em><em>)</em>

<em>Hope </em><em>I</em><em> </em><em>am</em><em> </em><em>Correct </em><em>I</em><em> </em><em>didn't</em><em> </em><em>really</em><em> </em><em>understand</em><em> </em><em>your</em><em> </em><em>question</em><em> </em><em>well</em><em>.</em><em> </em><em>But</em><em> </em><em>Good</em><em> </em><em>Luck</em><em>:</em><em>)</em>

3 0
3 years ago
Read 2 more answers
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