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ycow [4]
3 years ago
5

Did I do it right? If I didn't, Can u please explain why and give me the correct answer please?

Mathematics
2 answers:
snow_tiger [21]3 years ago
6 0
Yes u did
5/24=0.21
And 6/24 x200=50
I think u right
balandron [24]3 years ago
3 0

Answer:

Step-by-step explanation:

1) find the total of candies

3 + 6 + 5 + 2 + 8 = 24

2) divide the number of yellow candies  for the total amount of candies

5/24 = 0.21 (correct)

3) use a proportion

6 : 24 = x : 200

x = (200 *6) / 24 = 50 (correct)

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They will meet after four hours

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The boat moving at 2.5 mph will have travelled 10 miles after 4 hrs, and the boat moving at 5 mph will have travelled 20 miles after 4 hrs

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What’s the correct answer
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There's a strong negative association between the two variables.


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Describe the transformation of the graph of the parent function y= sqrt x+7 + 5​
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The function moves 7 to the left and 5 units up

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y = (x + 7) + 5

I think that's the problem you put.

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Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ
Allisa [31]

Answer:

Probability of stopping the machine when X < 9 is 0.0002

Probability of stopping the machine when X < 10 is 0.0013

Probability of stopping the machine when X < 11 is 0.0082

Probability of stopping the machine when X < 12 is 0.0399

Step-by-step explanation:

There is a random binomial variable X that represents the number of units come off the line within product specifications in a review of n Bernoulli-type trials with probability of success 0.91. Therefore, the model is {15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}. So:

P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002

P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013

P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082

P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399

Probability of stopping the machine when X < 9 is 0.0002

Probability of stopping the machine when X < 10 is 0.0013

Probability of stopping the machine when X < 11 is 0.0082

Probability of stopping the machine when X < 12 is 0.0399

8 0
3 years ago
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