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3 years ago

- What is the percent of increase from 5,000 to 8.000?

Mathematics

1 answer:

lana [24]3 years ago

4 0

Answer:

increase is 37.5%

Step-by-step explanation:

3000/8000 *100= 37.5%

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for every 5 coins carmen gets, she gives 2 to her brother Frankie. If carmen has 9 coins, how many does frankie have?

aliya0001 [1]

Since Carmen gives two coins to Frankie for every five that she gets, she effectively only gains 3 coins at a time. Therefore, if she has 9 coins, Frankie must have 6.

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4 years ago

Please answer A B C or D:

ss7ja [257]

C I think that's it idk I might be wrong sorry if I am

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3 years ago

Which is a solution to the equation?<br> (х-2)(х + 5) = 18?

jok3333 [9.3K]

$\bf (x-2)(x+5)=18\implies \stackrel{\mathbb{F~O~I~L}}{x^2+3x-10}=18\implies x^2+3x-28=0 \\\\\\ (x-4)(x+7)=0\implies x= \begin{cases} 4\\ -7 \end{cases}$

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3 years ago

I need help please asap

lana66690 [7]

Answer:

Part B

How does this change in setting affect Miguel? Choose two answers.

Miguel asks his mom if he can transfer to a different middle school.

Miguel becomes a more focused student.

Miguel begins to struggle in his Language Arts class as well.

Miguel makes a new friend.

Step-by-step explanation:

6 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