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Charra [1.4K]
2 years ago
13

Rhombus ABCD has a diagonal BD¯¯¯¯¯.

Mathematics
1 answer:
Alexxandr [17]2 years ago
7 0
Use gauthmath it can help with any math problem 3EMB3R use this code to get more tickets
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What is the slope of the line given by the equation below? y = - 3x + 5
ladessa [460]

Answer:

-3

Step-by-step explanation:

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2 years ago
Determine the equatiln of the graph and select the correct answer below. <br><br>​
olga_2 [115]

Answer:

C

Step-by-step explanation:

okay, this is actually pretty easy because it just works with simple translations.

(x+a)^2 + b

if a is positive, the graph shifts left

if a is negative, the graph shifts right

if b is positive, the graph shifts up

if b is negative, the graph shifts down

so since the quadratic, is starts at the origin and is shift 3 to the left, and 2 up:

the equation is

(x+3)^2 + 2

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3 years ago
Plz help me im begging u to help
Vedmedyk [2.9K]

Answer:

356 miles a day nkw help me

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3 years ago
Read 2 more answers
A spherical ball just fits inside a cylindrical can that is 10 centimeters tall, with a diameter of 10 centimeters. Which expres
fgiga [73]

Step-by-step explanation:

Volume:

volume \: of \:  a\: sphere =  \frac{4}{3} \pi {r}^{3}  \\  =  \frac{4}{3} \pi( {5)}^{3}

3 0
2 years ago
Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea
elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) =  P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

  • The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
  • There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways to select exactly 3 aces is:

4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052

5 0
3 years ago
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