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Anvisha [2.4K]
3 years ago
13

A jar contains 17 orange,7 pink, and 4 black marbles. A marble is drawn at random.What is the probability that the marble is pin

k or orange
Mathematics
1 answer:
Grace [21]3 years ago
5 0
Sample space = 17 + 7 + 4 = 28
P(pink) = 7/28
P(orange) = 17/28
P(pink or orange) = P(pink) + P(orange)
P(pink or orange) = 7/28 + 17/28
P(pink or orange) = 24/28
P(pink or orange) = 6/7 or 0.857143 or 85.7%
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Step-by-step explanation:


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2 years ago
What is the equation of the line that passes through the point (3,-1) and is perpendicular to the equation y= -3x+2?
Sauron [17]

Answer:

i think it is b or d im not for sure though

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Triangle ABC is dilated to produce triangle A'B'C' with scale factor 3/4. which describes the relationship between the two trian
Nadya [2.5K]

Answer:

ΔA'B'C' is a reduction of ΔABC and ΔA'B'C' is similar to ΔABC.

Step-by-step explanation:

It is given that the triangle ABC is dilated to produce triangle A'B'C' with scale factor 3/4.

If a figure is dilated then preimage and image are similar.

If scale factor is between 0 to 1, then preimage is reduction of image.

If scale factor is more that 1, then preimage is enlargement of image.

If scale factor is 1, then preimage is congruent to the image.

We know that

\frac{3}{4}=0.75

So,

0

Therefore, the ΔA'B'C' is a reduction of ΔABC and ΔA'B'C' is similar to ΔABC.

6 0
3 years ago
WILL GIVE BRAINLIEST FOR BEST ANSWER I NEED THIS NOW! THANK YOU!
Neporo4naja [7]
What we know:
Football field is 100 yards in length
End zones are 10 yards each in length
Perimeter between pylons is 306 2/3 yards

What we need to find:
a. Perimeter and area of one end zone
b. Perimeter and area of without end zones
c. Perimeter and area of playing field with end zones

First we need to find the measurements of the field between pylons using the perimeter of 306 2/3 yards. We already know the length is 100 yards so we need to find width (w).

P=2l + 2w
306 2/3=2 (100) + 2w
306 2/3= 200 + 2w
300 2/3-200=200-200+2w
106 2/3=2w
106 2/3/2=2/2w
53 1/3=w

a. Perimeter=2 (10)+2 (53 1/3)=126 2/3 yards
Area=10×53 1/3=533 1/3 yd²
b. Perimeter=2 (100)+2 (53 1/3)=306 2/3 yards
Area=100×53 1/3=5333 1/3 yd²
c. Perimeter=2 (120) + 2 (53 1/3)=346 2/3 yards
Area=120×53 1/3=6400 yd²
6 0
3 years ago
. A hat contains 100 coins, where 99 are fair but one is double-headed (always landing Heads). A coin is chosen uniformly at ran
SOVA2 [1]

Answer:0.565

Step-by-step explanation:

Hat contain 100 coins out of which 99 is fair and one is double headed.

Probability of getting 7 heads is sum of Head through fair coin+heads through biased coin

Probability of getting a head through a fair coin=0.5

P(7 heads)=0.99\times 0.5^7+0.01\times 1

P(Double head |7 heads )=\frac{Probability\ of\ 7\ heads\ and\ double\ headed}{probability\ of\ 7\ heads}

P(Double head |7 heads )=\frac{0.01\times 1}{0.99\times 0.5^7+0.01\times 1}

=\frac{0.01}{0.99\times 0.0078+0.01}=0.565

8 0
3 years ago
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