All categories

3 years ago

Select three ratios that are equivalent to 8:20 Choose 3 answers: (A. 1:4

Mathematics

1 answer:

denpristay [2]3 years ago

8 0

Answer:

8:20

divide by 2

B. 4:10

multiply by 3

C. 24: 60

divide by 4

D. 2:5

You might be interested in

Carl Brownstein is a golf course groundkeeper whose annual gross pay is $45,500. He has exemptions of $3,000. The tax state on t

zhenek [66]

Answer:

Step-by-step explanation:

7 0

3 years ago

Drag each label to the correct location on the table. Each label can be used more than once, but not all labels will be used.

Natalka [10]

Answer:

what lable i dont see nothing

Step-by-step explanation:

7 0

3 years ago

What is the explicit rule for the sequence?<br><br> 10.5, 9, 7.5, 6, 4.5, 3, ...

makvit [3.9K]

Answer: the rule is to add 1.5

3 + 1.5 = 4.5
4.5 + 1.5 = 6
6 + 1.5 = 7.5
etc...

6 0

3 years ago

Tom is putting money into a savings account. He starts with $650 in the savings account, and each week he adds $60 . Let S repre

mario62 [17]

S = 60W + 650 <== ur equation

after 18 weeks...so sub in 18 for W

S = 60(18) + 650
S = 1080 + 650
S = 1730 <==

7 0

4 years ago

Read 2 more answers

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