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3 years ago

Twice the sum of a number and 5 equals 9 as a equation

Mathematics

1 answer:

ziro4ka [17]3 years ago

7 0

Hello!

I believe it looks like this..

2*x+5=9

Btw, x = 3 is your answer if you're wondering..

Hope this helps! ☺♥

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1.morgan is 15 years old younger than Mrs.santtos their combined age is 442. jenny won the ping-pong championship eight more tim

swat32

If Morgan is 15 years younger then Mrs.Santos and there combined age is 442 that means that Mrs.Santos is 221, and Morgan is 206.

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3 years ago

If you roll a number cube 12 times, about how many times would you expect to roll a 5 or 6

Marrrta [24]

You have a 1/3 chance each time you role So 1/3 x 12 = 4 times that you would role a 5 or a 6

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2 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

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3 years ago

Selmas class making a care package to give to victims of a natural disaster.Selma packs one box in 5 minutes and has already pac

bixtya [17]

Each of them need to work 105 minutes more in order to have packed the same number of boxes.

Explanation

Suppose, they need to work for $x$ minutes more in order to have packed the same number of boxes.

Selma packs one box in 5 minutes and Trudy packs one box in 7 minutes.

So, the number of boxes packed by Selma in that $x$ minutes $=\frac{x}{5}$ and the number of boxes packed by Trudy in $x$ minutes $=\frac{x}{7}$

Given that, Selma and Trudy have already packed 12 and 18 boxes.

Now if each of them packed the same number of boxes, then the equation will be......

$12+\frac{x}{5}=18+\frac{x}{7}\\ \\ \frac{x}{5}-\frac{x}{7}= 18-12\\ \\ \frac{7x-5x}{35}=6\\ \\ \frac{2x}{35}=6\\ \\ 2x= 210\\ \\ x= \frac{210}{2}=105$

So, each of them need to work 105 minutes more in order to have packed the same number of boxes.

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3 years ago

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Andre45 [30]

2 ones and 8 hundredths

00.00
the first zero is tens, ones, and after the decimal, to the right is the tenths, hundredths and so on

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3 years ago