What is the answer for x

Help me please this is donr tomorrow and if i don't finish it my teacher will be so mad

Genrish500 [490]

3/4 since (9 1/2 = length* width with makes the base) and divide it by 7 1/8.<span />

4 0

3 years ago

PLEASE HELP WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!

Paraphin [41]

the overhead, fixed costs, is $1000.

the variable cost is $200, so if she sells "x" tiles, then her costs is 200*x or 200x.

total costs then will just be C(x) = 200x + 1000.

namely both costs added together.

Revenue is just how much it's been taken in from the sales, since each tile is selling for $240, then the revenue is simply 240*x or just R(x) = 240x.

break even point is when those two guys equal each other

$\bf \stackrel{R(x)}{240x}=\stackrel{C(x)}{200x+1000}\implies 40x=1000\implies x=\cfrac{1000}{40}\implies x=25$

what if the cost for each tile were $220, 20 bucks more?

$\bf \stackrel{R(x)}{240x}=\stackrel{C(x)}{220x+1000}\implies 20x=1000\implies x=\cfrac{1000}{20}\implies x=50$

well, if the cost for each tile is $200, her break-even point is at 25 units, namely once she has sold 25 tiles, she's has lost nothing, has won nothing either, but hasn't lost anything.

if the cost is however $220 instead, her break-even point comes much later, at 50 units, so she'll have to sell more tiles in order to not have any losses, so she's worse off if the cost is more or course.

4 0

3 years ago

Jonathon and Raymond earn a base salary plus commission. Jonathon earns $1500 plus 10% of his sales each month. Raymond earns $1

joja [24]

Answer:

The Jonathon and Raymond need to sell 6000 sales to earn the same amount each month.

Step-by-step explanation:

Given data

Let no. of sales = x

Jonathon earning = $ 1500 + $\frac{10}{100} x$

Raymond earning = $ 1200 + $\frac{15}{100} x$

Given that the earnings of Jonathon & Raymond are same.

1500 + $\frac{10}{100} x$ = 1200 + $\frac{15}{100} x$

300 = $\frac{5}{100} x$

x = 6000 sales

Therefore the Jonathon and Raymond need to sell 6000 sales to earn the same amount each month.

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:

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