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bekas [8.4K]
2 years ago
7

Can someone help me for once ?​

Mathematics
1 answer:
hichkok12 [17]2 years ago
6 0

Answer:

The answer is A)-4,6

Step-by-step explanation:

sorry if wrong

hope this helps

plz give brainliest

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An online furniture store sells chairs for $50 each and tables for $550 each. Every day, the store can ship a maximum of 32 piec
Hunter-Best [27]

Answer:

you have to sell 6 tables to meet all requirements

Step-by-step explanation:

chairs=$50x

tables=$550x

24 chairs×50= $1200

4100-1200= $2900

take $2900 and divide by $550 to find the exact number of tables which is 5 but selling 5 tables and 24 chairs doesnt reach the $4100 mark so I rounded up to 6 tables which doesnt surpass the maximum number of furniture(32) but beats the $4100 mark

3 0
3 years ago
An urn contains nine red and one blue balls. A second urn contains one red and five blue balls. One ball is removed from each ur
il63 [147K]

Answer:

0.362

Step-by-step explanation:

When drawing randomly from the 1st and 2nd urn, 4 case scenarios may happen:

- Red ball is drawn from the 1st urn with a probability of 9/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this case to happen is (9/10)*(1/6) = 9/60 = 3/20 or 0.15. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 5 blue)/(8 red + 1 blue + 5 blue) = 6/14 = 3/7.

- Red ball is drawn from the 1st urn with a probability of 9/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (9/10)*(5/6) = 45/60 = 3/4 or 0.75. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 4 blue)/(8 red + 1 blue + 1 red + 4 blue) = 5/14

- Blue ball is drawn from the 1st urn with a probability of 1/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (1/10)*(5/6) = 5/60 = 1/12. The probability that a ball drawn randomly from the third urn is blue given this scenario is (4 blue)/(9 red + 1 red + 4 blue) = 4/14 = 2/7

- Blue ball is drawn from the 1st urn with a probability of 1/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this event to happen is (1/10)*(1/6) = 1/60. The probability that a ball drawn randomly from the third urn is blue given this scenario is (5 blue)/(9 red + 5 blue) = 5/14.

Overall, the total probability that a ball drawn randomly from the third urn is blue is the sum of product of each scenario to happen with their respective given probability

P = 0.15(3/7) + 0.75(5/14) + (1/12)*(2/7) + (1/60)*(5/14) = 0.362

8 0
3 years ago
Here are the questions:
yaroslaw [1]

Answer:

Q1) 0.46 Q2)0.2333

Step-by-step explanation:

Q1) P=(4+12+7)/50 =0.46

Q2) P=7/(4+7+10+9)

6 0
3 years ago
Use the parabola tool to graph the quadratic function
shepuryov [24]

Your parabola should look like this image. The vertex is at (-1, 9), and you can use any of the x-intercepts (-4, 0) & (2, 0) or the y-intercept (0, 8) as a second point.

6 0
3 years ago
Al and Bill each have two bullets. They fire alternately at a glass bottle. Al has hitting probability 1/3 and Bill 1/4. What is
enot [183]
P(Al hits bottle first time) = 1/3
P(Al misses the first shot but hits on his second shot) =  
P( Al misses and bill misses and Al hits) = 2/3 * 3/4 * 1/3 = 1/6

So required probability = 1/3 + 1/6 =  1/2
8 0
3 years ago
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