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3 years ago

The are 85 girls 29 boys and 123 people in the class in total how many adults are in the classroom??

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

7 0

85+29=114
123-114= 9
9 adults were in the room.

Alex_Xolod [135]3 years ago

3 0

9 adults

Step-by-step explanation:

85+29=114

123-114=9

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Which method can be used to prove that AJKL is congruent to AJML? SSS OHL SAS ASA

jeka57 [31]

Answer:

I think the answer is SAS(side angle side).

Step-by-step explanation:

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3 years ago

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A bridge hand is made up of 13 cards from a deck of 52. Find the probability that a hand chosen at random contains at least .

Tema [17]

Answer:

10%

Step-by-step explanation:

Since the question is not complete, I will be solving for probability of 3 9s

To solve this, we would be using the principle of combination, and thus

A deck of 52 hours has 4 nines, the probability that it contains at least 3 9s is mathematically represented by the equation.

P ( x ≥ 3 ) = ( 48 C 10 · 4 C 3 ) / ( 52 C 13 ) + ( 48 C 9 · 4 C 4 ) / ( 52 C 13 )

Remember I told us were going to be using the principle of combination. Well then

nCr = n!/r!(n - r)!

Now, applying the above stated formula to the equation, we have

48C10 = 48!/10!(38!)

4C3 = 4!/3!(1!)

52C13 = 52!/13!(39!)

48C9 = 48!/9!(39!)

4C4 = 4!/4!(0!)

Solving for each independently, we find that

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3 years ago

Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab

Sphinxa [80]

Answer: a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Step-by-step explanation:

Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:

Since we have given that

$\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5$

Using the poisson process, we get that

(a) less than or equal to 12

probability= $P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980$

(b) equal to 13

probability= $P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013$

probability= $P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020$

(d) equal to 20

probability= $P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026$

(e) between 10 and 15, inclusively

probability= $P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318$

Hence, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

6 0

3 years ago

HELPPP i need to find standard form and general form, please help

Rudiy27

The equation in slope-intercept form is y = 13/8x -17/8 and the equation in standard form is 13x -8y = 17

<h3>Equation of a line</h3>

A line is the shortest distance between two points. The equation of a line in slope-intercept form is y = mx + b while as a standard notation is Ax+By = C

Given the coordinate points (-3, -7) and (5, 6)

Determine the slope

Slope = 6-(-7)/5-(-3)

Slope = 13/8

For the y-intercept

6 = 13/8(5) + b

6 = 65/8 + b

b = 6 - 65/8

b = -17/8

The equation in slope-intercept form is y = 13/8x -17/8

Write in standard form

Recall y = 13/8x - 17/8

Multiply through by 8

8y = 13x - 17

-13x + 8y = -17

13x -8y = 17

Hence the equation in standard form is 13x -8y = 17

Learn more on equation of a line here: brainly.com/question/13763238

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2 years ago

Find the first 5 terms in the following sequence: a1 = 23, a, an-1 + 7, n > 2

sasho [114]

Answer:

Step-by-step explanation:

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3 years ago