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dimaraw [331]
3 years ago
13

A total of 636 tickets were sold for the school play. They were either adult tickets or student tickets. There were 64 fewer stu

dent tickets sold than adult tickets. How many adult tickets were sold?
Mathematics
1 answer:
NARA [144]3 years ago
7 0
Five hudred sixty two. 562
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8 times the difference between a number and 6 is equal to 4 times the number...
astraxan [27]

Answer:

(A) 12

Step-by-step explanation:

First, lets define the number as x.

"8 times the difference between a number and 6 is equal to 4 times the number" is translated into "8(x - 6) = 4x"

8(x - 6) = 4x

8x - 48 = 4x

4x = 48

x = 12

6 0
3 years ago
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Use the order of<br> operations to evaluate<br> the expression<br> (-8X14)-(-2X-10)
Leviafan [203]

Answer:

the order of operations is .called BODMAS rule

B.brackets

O.of

D.division

M.multiplication

A.addition

S.subtraction

Step-by-step explanation:

= (-112)-(20)

= -132

6 0
3 years ago
Please help me!!!!!!!!!!
True [87]

Answer:

it think it is D

Step-by-step explanation:

just take one of the number and move it for example 4x right and 2y up and then look at where the connection is.

6 0
3 years ago
‼️‼️‼️HELP HELP ASAP PLEASE AND THANK YOU AND DONT FORGET TO SHOW WORK ILL MARK YOU THE BRAINLIEST ‼️‼️‼️​
solniwko [45]

Answer:

1809.557368

Step-by-step explanation:

5 0
3 years ago
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Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
ludmilkaskok [199]

Answer:

\lambda \geq 6.63835

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99

And now we can solve for \lambda

0.01 \geq e^{-\lambda}(1+\lambda)

Applying natural log on both sides we have:

ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)

ln(0.01) \geq -\lambda+ln(1+\lambda)

\lambda-ln(1+\lambda)+ln(0.01) \geq 0

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
3 years ago
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