Nine out of ten students prefer math class over lunch. How many students do not prefer math if 200 students were asked?

1 answer:

3 0

So, 9 out of 10 students prefer class, so, obviously, it would not be 100. It would be 150. So in conclusion, 50 students prefer lunch over math class, whilst 150 other students prefer math class over lunch.

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What is the perimeter of a triangle with vertices of (5,1) (-3,3) (-7,-3)

timurjin [86]

$Vertices:\\
A(5,1)\\B(-3,3)\\C(-7,-3)\\\\ You\ must\ find\ length\ of\ AB,\ BC,\ AC.\\\\
A(5,1)\ \ \ \ \ \ B(-3,3)\\\\Distance\ between\ A \ and\ B:\\AB=\sqrt{(x_B-x_A)^2+(y_A-y_B)^2}\\\\AB=\sqrt{(-3-5)^2+(3-1)^2}\\AB=\sqrt{8^2+2^2}\\AB=\sqrt{64+4}\\AB=\sqrt{68}
$ $
 B(-3,3)\ \ \ \ C(-7,-3)\\\\Distance\ between\ B \ and\ C:\\BC=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}\\\\BC=\sqrt{(-7-(-3))^2+(-3-3)^2}\\BC=\sqrt{(-4)^2+(-6)^2}\\BC=\sqrt{16+36}\\BC=\sqrt{52}$ $A(5,1)\ \ \ \ \ \ C(-7,-3)\\\\Distance\ between\ A \ and\ C:\\AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}\\\\AC=\sqrt{(-7-5)^2+(-3-1)^2}\\AC=\sqrt{(-12)^2+(-4)^2}\\AC=\sqrt{144+16}\\AC=\sqrt{160}$ $Perimeter\ of\ triangle=\ AB+AC+BC=\\\sqrt{68}+\sqrt{52}+\sqrt{160}=2\sqrt{17}+2\sqrt{13}+2\sqrt{40}$

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3 years ago

What is the sum of 3b(b-4) and 6(b-8)

lora16 [44]

Answer:

This equals $3b^2-6b-48$

Step-by-step explanation:

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3 years ago

Henry says he can write an 8 in each box to make both equations true. Is he correct?

Rama09 [41]

Answer:

yes it's true because 3 groups of 8 is 24

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Question 3

DanielleElmas [232]

It would be -6. -4-6 is -10

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The Census Bureau says that the 10 most common names in the United States are (in order) Smith, Johnson, Williams, Jones, Brown,

scZoUnD [109]

Using binomial distribution where success is the appearing of any of the top 10 most common names, thus probability of success (p) is 9.6% = 0.096 and the probability of failure = 1 - 0.096 = 0.904. Number of trials is 11.
Binomial distribution probability is given by P(x) = nCx (p)^x (q)^(n - x)
Probability that none of the top 10 most common names appears is P(0) = 11C0 (0.096)^0 (0.904)^(11 - 0) = (0.904)^11 = 0.3295
Thus, the probability that at least one of the 10 most common names appear is 1 - 0.3295 = 0.6705

Therefore, I will be supprised that none of the names of the authors were among the 10 most common names given that the probability that at least one of the names appear is 67%.

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3 years ago