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Ymorist [56]
3 years ago
5

Sean’s age and Helen’s age are in the ratio of 3:5. Two years ago, Helen was twice as old as Sean was then. Find their present a

ges.
Mathematics
1 answer:
PolarNik [594]3 years ago
3 0
Lets Sean's age be x
Helen's age by y
x/y=3/5
x=3/5y

Second equation
2(x-2)=y-2

on solving u get Helen as 10 and Sean as 6
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Answer:

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Noah bought 15 baseball cards for $9.00. Assuming that each baseball card costs the same amount, answer the following questions.
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Answer:

270 dollars

Step-by-step explanation:

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3 years ago
Solve the system of equations by any method.<br><br><br><br> 5x+9y =16<br> x+2y =4
Alisiya [41]

Answer:

The system of equations has one solution at (-4, 4).

Step-by-step explanation:

We are given the system of equations:

\displaystyle\left \{ {{5x+9y=16} \atop {x+2y=4}} \right.

We can use elimination to solve this system. We need to multiply the second equation by -5 so we can cancel out our x-terms.

-5\times(x+2y=4) \rightarrow -5x - 10y = -20

Therefore, our system now becomes:

\displaystyle\left \{ {{5x+9y=16} \atop {-5x-10y=-20}} \right.

Now, we can add these two equations together and solve for y.

\displaystyle(5x + 9y) + (-5x - 10y) = 0 - y\\\\16 + (-20) = -4\\\\-y = -4\\\\\frac{-y}{-1}=\frac{-4}{-1}\\\\y = 4

Now, we can substitute our value for y into one of the equations and solve for x.

x+2(4)=4\\\\x + 8 = 4\\\\x = -4

Therefore, our final solution is (-4, 4).

6 0
3 years ago
Translate to equation<br> Two times the number which is three less than X is eight
GrogVix [38]
I’m not sure, you’re not elaborating enough to this equation. Are you sure you’re not missing a phrase of some sort?
8 0
2 years ago
Read 2 more answers
There are several scenarios described below. For each of them, do the following (note: R.V. means random variable) (1) Define th
frutty [35]

Answer:

a) The Ohio Bureau of Motor Vehicles states that 7 out of 8 people pass the written driver’s test.

Let X be the number of test given by the test taker to pass out.

So X~Geometric(p) where p=Probability that for a particular test anyone will pass the written test=7/8

and here support of X be equal to 1,2,3,..... i.e X is a Natural number

So ,probability that he will pass the written test in fewer than 4 tries

=P(X<4)

=\sum _{x=0}^{3}p(1-p)^{x-1}

b) LAIMO manufacturing company makes parts for the auto industry. Approximately 3% of the parts it makes are defective.

So let X=number of non defective parts sampled before the 3rd defective part is sampled

then X~Negative Binomial(r,p) where here r=3 and p=Probability that a randomly selected part is defective= 0.03

where support of X is {0,1,2,3,...}

So the probability that the third defective part is the 20th one sampled.

P(X=20-3=17)

=\binom{r+16}{17}p^r(1-p)^{16}

c) A BigMart store is going to hire 3 new cashiers. It has 18 applicants (10 male, 8 female) for these 3 cashier jobs.

So let X be number of female cashier appointed.

Here X~Hypergeometric(3,8,18) where

f(x)=P(X=x)

=\left\{\begin{matrix} \frac{\binom{8}{x}.\binom{10}{3-x}}{\binom{18}{3}} & ,x=0,1,2,3\\ 0 & ,otherwise \end{matrix}\right.

So the probability that none of the positions are filled by females

=P(X=0)

d) A gardener is inspecting the fall flowers in her garden. She notices, on average, 4 bugs on a flower. She randomly picks one flower from her garden.

Let X be the numbers of bugs on that flower

So X follows Poisson distribution with mean 4 where support of X is {0,1,2,3.....}

So the probability that the flower she picked has at least one bug on it

=P(X\geq 1)=1-P(X=0)

=1-e^{-4}\frac{4^x}{x!}|_{x=0}

e) A student is taking a true/false test that consists of 15 questions. Based on past performance the student has approximately a 70% chance of getting any individual question correct.

So let X be the number of questions that are correct among those 15 questions.

so X~Binomial(n,p) where n=15 and p=Probability that he get an individual question correct =0.7

where support of X be {0,1,2,3,...,15}

So the probability that the student gets at least 60% of the questions on the test correct or 15x60%=9 questions are correct

=P(X\geq9)

=\sum _{x=9}^{15}\binom{15}{x}p^x(1-p)^{15-x}

f) A certain radio station’s phone lines are busy approximately 95% of the time when trying to call during a contest.

Let X denotes the number of calls to get into the contest.

So X~Geometric(p) where p=Probability that in a call I get through into the contest=1-0.95=0.05

support of x={1,2,3,....}

So the probability that the 4 th time you call is the 1st time you get through during a contest.

=P(X=4)

=p(1-p)^4

4 0
3 years ago
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