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liq [111]
3 years ago
8

El cociente de una división es nueve y el residuo es 20. Si el divisor se disminuye en tres, entonces el cociente aumenta en tre

s y el residuo disminuye en seis. ¿Cuál es el valor del divisor y del dividendo?
Mathematics
1 answer:
ludmilkaskok [199]3 years ago
6 0
Sorry i dont speak spanish.
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A store sells a package of 7 water bottles for $5.60.<br> What is the price per water bottle?
Luden [163]
Since there is 7 water bottles in a package and the total price is 5.60, you divide 7 by 5.60 which gives you 1.25 per bottle
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3 years ago
I need help please only help if you know the answer
vlada-n [284]

Answer:

D

Step-by-step explanation:

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a graphic designer wants to know the number of regions that are formed when circles overlap in particular. can she find a rule t
kumpel [21]

Answer:

A. 2

Step-by-step explanation:

Its successive multiples and it increases:

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it increases by 2 every time which is successive multiples so the answer is A, 2.

3 0
3 years ago
Dy/dx ln(xy)=e^(x+y)
Westkost [7]
In(xy) = e^(x+y)

(xy)'/xy  = (x+y)' e^(x+y)

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4 0
3 years ago
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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