Beth is solving this equation: 1/x + 3= 3/x

6. Tasha walked 6,618 feet. What is the distance she walked in yards?

juin [17]

The answer is 2206.

Step-by-step explanation:

3 feet = 1 yard

so you divide 6618 by 3 to get 2206

6 0

3 years ago

Will ran the diagonal distance across a square field measuring 40 yards on each side. James ran the diagonal distance across a r

slamgirl [31]

Will ran the longest

He ran for 56.57 yards, 13.56 yards longer than James.

Step-by-step explanation:

Step 1 :

Will ran the diagonal across a square field measuring 40 yards in each side.

The diagonal of a square can be obtained by the square root of the sum of the squares of its 2 sides [Because it forms the hypotenuse of a right angle triangle]

Hence when the side is 40 yards , the diagonal would be

$\sqrt{40^{2} + 40^{2}} = \sqrt{1600 + 1600} = \sqrt{3200} =56.57 yards$

So Will ran for 56.57 yards

Step 2 :

James ran the diagonal of a rectangular field with 25 yards length and 35 yards width.

The diagonal of the rectangle can be obtained by the square root of the sum of squares of its length and width.

Hence when the length is 25 yards and width is 35 , the diagonal would be

$\sqrt{35^{2} + 25^{2}} = \sqrt{1225 + 625}} = \sqrt{1850} = 43.01$ yards

So James ran for 43.01 yards

Step 3 :

Will ran for 56.57 yards and 43.01 yards.

Hence Will ran for longer distance of 56.57 yards, which is 13.56 yards more than James.

5 0

3 years ago

Use the following data set and a separate piece of paper to make a line plot that will help you to answer questions 2-7. (6, 4,

Montano1993 [528]

Hello? If you don't know, mode means the most occurring number. And in the data, the number 6 has occurred four times. Which means that the answer is 6.

3 0

3 years ago

Read 2 more answers

The cost y to rent a tent is proportional to x days. if it costs $56 to rent a tent for 7 days what is the formula

Alona [7]

Probably y = 8x. 56 = 8(7), so that works. Without more data, I couldn't say if it could be something else, so that's what I go with.

4 0

3 years ago

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