Ask question

Ask question

All categories

4 years ago

12

Susie's sushi place two orders with its fish supplier one order was for 15 lb of salmon and 8 lb of tuna the order total $228 th

e other order was for 12 lb of salmon and 6 lb of tuna this ordered total for 180 what is the cost for 1 lb of salmon in 1 lb of tuna

1 answer:

VashaNatasha [74]4 years ago

7 0

30 is the answer

Jejhdjsjs

You might be interested in

Section 1 - Question 1

Morgarella [4.7K]

Answer:

a and bwhich step did he make his first error simplifying the expression?

A

Step

3 0

3 years ago

5/6 + 4/6<br><br>a) 1/6<br>b) 9/12<br>c)1 3/6<br>d) 1 2/6

babymother [125]

Answer:

this is the answer Hope's it helps

7 0

3 years ago

Kishore bought a basket containing 30 orangs for rs 310. If he sells the orangs at rs 12 each , what Would be it's profit or los

iris [78.8K]

Given : Kishore bought a basket containing 30 oranges for rs 310. If he sells the oranges at rs 12 each , what Would be it's profit or loss?

Solution :

Cost price of oranges = rs.310

As we know that each oranges cost rs.12

So, selling price of oranges is 30 × 12 = rs.360

Now, <u>finding profit or loss</u>

Profit = Selling Price - Cost Price

Profit = rs.360 - rs.310

Profit = rs.50

So, the profit is ₹50

7 0

3 years ago

Which situation is most likely to have a constant rate of change?

Natasha_Volkova [10]

Answer: B.

HOPE THIS HELPS :D

5 0

4 years ago

The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.What is

Lesechka [4]

Answer:

a) The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.

b) The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.

Step-by-step explanation:

Given : The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.

To find : What is the probability that

(a) the total gross sales over the next 2 weeks exceeds $5000;

(b) weekly sales exceed $2000 in at least 2 of the next 3 weeks? What independence assumptions have you made?

Solution :

Let $X_1$ and $X_2$ denote the sales during week 1 and 2 respectively.

a) Let $X=X_1+X_2$

Assuming that $X_1$ and $X_2$ follows same distribution with same mean and deviation.

$E(X)=E(X_1+X_2)=E(X_1)+E(X_2)$

$E(X)=2\mu = 2(220)=4400$

$\sigma_X=\sqrt{var(X_1+X_2)}$

$\sigma_X=\sqrt{2\sigma^2}$

$\sigma_X=\sqrt{2}\sigma$

$\sigma_X=230\sqrt{2}$

So, $X\sim N(4400,230\sqrt{2})$

$P(X>5000)=1-P(X\leq5000)$

$P(X>5000)=1-P(Z\leq\frac{5000-4400}{230\sqrt{2}})$

$P(X>5000)=1-P(Z\leq1.844)$

$P(X>5000)=1-0.967$

$P(X>5000)=0.0321$

The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.

b) The probability that sales exceed teh 2000 and amount in at least 2 and 3 next week.

We use binomial distribution with n=3.

$P(X>2000)=1-P(X\leq2000)$

$P(X>2000)=1-P(Z\leq\frac{2000-2200}{230})$

$P(X>2000)=1-P(Z\leq-0.87)$

$P(X>2000)=1-0.1922$

$P(X>2000)=0.808$

Let Y be the number of weeks in which sales exceed 2000.

Now, $P(Y\geq 2)$

So, $P(Y\geq 2)=P(Y=2)+P(Y=3)$

$P(Y\geq 2)=^3C_2(0.8077)^2\cdot (1-0.8077)+^3C_3(0.8077)^3$

$P(Y\geq 2)=0.37635+0.52692$

$P(Y\geq 2)=0.90327$

The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.

3 0

3 years ago