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3 years ago

I need help with this problemwill mark as brainliest

Mathematics

1 answer:

Arada [10]3 years ago

7 0

Answer:

The shop can cut the cost of production per sandwich to $3.75. The revenue function would then be defined as R(x)=9x.

Step-by-step explanation:

We want to make as big profit as possible by making as little sandwiches as possible. This means that we want the profit, which is $revenue- production\: cost$ to be as big as possible.

For the first choice the profit per sandwich is $8.25-4=4.25$ dollars.

For the second choice the profit per sandwich is $9-3.75=5.25$ dollars.

For the third choice the profit per sandwich is $8.5-3.5=5$ dollars.

We see here that for the second choice the profit is greatest, therefore this choice is most suitably correct.

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A team manager buys a team jacket and a team cup for each of the 10 players on the dodgeball team. Each team cup costs $10.70. T

Ilya [14]

Answer:

293/10=u

thats the equations

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2 years ago

A load of bricks is twice as heavy as a load of sticks. The total weight of 4 loads of bricks and 4 loads of sticks is 771 kilog

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It's 501.15 kg

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3 years ago

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lubasha [3.4K]

$80.00 hope this helps!

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3 years ago

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

WILL MARK BRAINLIEST

Liula [17]

Answer:

possible numbers, he can make with cards 5 9 0 :

5, 9, 0

5, 0, 9

9, 5, 0

9, 0, 5

0, 5, 9

0, 9, 5

5 0

2 years ago

I need help with this problem​will mark as brainliest

I need help with this problemwill mark as brainliest