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3 years ago

13

how many fewer rainy days are there than sunny and cloudy days when there are 6 sunny days, 4 cloudy days and 7 rainy days

1 answer:

Cloud [144]3 years ago

8 0

3 more fewer days because 6+4=10 and 10-7=3.

Hope that helped:)

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Compare the following numbers 52,458 blank 52,485

skelet666 [1.2K]

52,458<52,485
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3 years ago

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Ok so Ive been having trouble with Math can anyone help??

Illusion [34]

For the first bit, put $930 + 7.5x and the second part is 58,000, 930 x 7.5 x 8

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3 years ago

1. Function f is transformed into function g by shifting the graph 4 units to the left and 8 units up.

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3 years ago

Daniel worked for 2 1/3 hours and earned $18.20. How much would he make if he worked for 5 1/4 hours? Enter your answer in the b

Aleks04 [339]

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2 years ago

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A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple

tester [92]

Answer: The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation: We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

$n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.$

Therefore, the probability of event A is given by

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.$

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

4 0

2 years ago