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tresset_1 [31]
3 years ago
9

Find the distance between (–3, –5) and (3, 4).

Mathematics
1 answer:
zhenek [66]3 years ago
4 0

Answer:(6,9)

Step-by-step explanation:

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HELPP PLEASE ASAP Use the Pythagorean theorem to find the missing length in the right triangle
qwelly [4]

Answer:

c = 10

Step-by-step explanation:

The pythagoras theorem :

a² + b² = c²

a = 6 and b = 8

6² + 8² = c²

36+64 = c²

100 = c²

c = √100

c= 10

Hope this helped and brainliest please

3 0
2 years ago
What is the surface area of a sphere with a radius of 19 units?
almond37 [142]

Hope this helps. Please mark brainliest if it did

Answer: 4536.46

5 0
3 years ago
Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five event
mezya [45]

Answer:

The probability of Javier volunteers for less than three a month is 25%

Step-by-step explanation:

According to the questions,

The probability of Javier to attend exactly 5 events :-

P ( x = 5 ) = 25% = \frac{1}{4}

The probability of Javier to attend exactly 4 events :-

P ( x = 4 ) = \frac{1}{4}

The probability of Javier to attend exactly 3 events :-

P ( x = 3 ) = \frac{1}{4}

The probability of Javier to attend exactly 2 events :-

P ( x = 2 ) = \frac{3}{20}

The probability of Javier to attend exactly 1 events :-

P ( x = 1 ) = \frac{1}{20}

The probability of Javier to attend no events :-

P ( x = 0 ) = \frac{1}{20}

Now we have to calculate the probability of Javier volunteers for less than three P ( x < 3 ).

P ( x < 3 ) = P ( x = 0 ) + P ( x = 1 ) + P ( x = 2 )

P ( x < 3 ) = \frac{1}{20} + \frac{1}{20} + \frac{3}{20}   = \frac{1 + 1 + 3}{20}  = \frac{5}{20}  = \frac{1}{4}   = 0.25 = 25 %.

8 0
3 years ago
Number sense which subtraction  sentences show you how to find 15- 7
Mars2501 [29]
You can use addition to solve this problem. 7+_= 15
6 0
3 years ago
A coin is tossed 5 times. Find the probability that all are heads. Find the probability that at most 2 are heads.
fenix001 [56]

Answer:

1/32

15/32

Step-by-step explanation:

For a fair sided coin,

Probability of heads, P(H) = 1/2

Probability of tails P(T)  = 1/2

For a coin tossed 5 times,

P( All heads)

= P(HHHHH),

= P (H) x P(H) x P(H) x P(H) x P(H)

= (1/2) x (1/2) x (1/2) x (1/2) x (1/2)

= 1/32 (Ans)

For part B, it is easier to just list the possible outcomes for

"at most 2 heads" aka "could be 1 head" or "could be 2 heads"

"One Head" Outcomes:

P(HTTTT), P(THTTT) P(TTHTT), P(TTTHT), P(TTTTH)

"2 Heads" Outcomes:

P(HHTTT), P(HTHTT), P(HTTHT), P(HTTTH), P(THHTT), P(THTHT), P(THTTH), P(TTHHT), P(TTHTH), P(TTTHH)

If we count all the possible outcomes, we get 15 possible outcomes representing "at most 2 heads)

we know that each outcome has a probability of 1/32

hence 15 outcomes for "at most 2 heads" have a probability of

(1/32) x 15  = 15/32

8 0
3 years ago
Read 2 more answers
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