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3 years ago

Greatest to least 0.12,0.1225,3/25,7/125

Mathematics

2 answers:

nadezda [96]3 years ago

3 0

.1225,.12, 7/125 the other fraction equals .12

Lubov Fominskaja [6]3 years ago

3 0

Answer:

0.1225, 3/25, 0.12, 7/125

Step-by-step explanation:

To solve this, you need to decide if you want decimals or fractions and get all of them the same.

0.12 = 3/25

0.1225 = 49/400

3/25 = 0.12

7/125 = 0.056

On decimals from the greatest to the least:

0.1225, 0.12, 0.12, 0.056

On fractions from the greatest to the least:

49/400, 3/25, 3/25, 7/125

On the given ones from the greatest to the least.

0.1225, 3/25, 0.12, 7/125

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Allisa [31]

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Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

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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

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