What is seven hundred thirty-one million, nine hundred thirty-four thousand, thirty in standard form

Mathematics

2 answers:

GrogVix [38]3 years ago

7 0

731,934,013
You can just list the numbers while you read and put them in order then you have your answer

Sindrei [870]3 years ago

6 0

731,934,030

please mark the brainiest

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I am confused on how to do this? please help!

IceJOKER [234]

You would take random numbers preferable smaller ones such as -1,0,1,2
You would put those in the place of x
Then you would square the value of x. After that you subtract 4. Whatever you come out with is your y value. Then you put the x value on the x axis and from that point count up or down to the number you need on the y axis. Together that would be your point

6 0

3 years ago

In a survey of adults who follow more than one sport, 30% listed football as their favorite sport. You survey 15 adults who foll

maxonik [38]

Answer: 0.5

Step-by-step explanation:

Binomial probability formula :-

$P(x)=^nC_x\ p^x(q)^{n-x}$ , where P(x) is the probability of getting success in x trials , n is the total trials and p is the probability of getting success in each trial.

Given : The probability that the adults follow more than one game = 0.30

Then , q= 1-p = 1-0.30=0.70

The number of adults surveyed : n= 15

Let X be represents the adults who follow more than one sport.

Then , the probability that fewer than 4 of them will say that football is their favorite sport,

$P(X\leq4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)\\\\=^{15}C_{0}\ (0.30)^0(0.70)^{15}+^{15}C_{1}\ (0.30)^1(0.70)^{14}+^{15}C_{2}\ (0.30)^2(0.70)^{13}+^{15}C_{3}\ (0.30)^3(0.70)^{12}+^{15}C_{4}\ (0.30)^4(0.70)^{11}\\\\=(0.30)^0(0.70)^{15}+15(0.30)^1(0.70)^{14}+105(0.30)^2(0.70)^{13}+455(0.30)^3(0.70)^{12}+1365(0.30)^4(0.70)^{11}\\\\=0.515491059227\approx0.5$