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4 years ago

Simplify this expression 2\3(-9m+ 12)

1 answer:

5 0

So with the 2/3 you factor out the 3 from the expression by multiplying 3, which cancels out the bottom number 3 and leaves you with 2(-3m+4). You then distribute 2 through the parenthesis which gives you the answer -6m+8 <span />

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In the first session 40 student attended.Of these 40 students 60% were girls. How many girls attended the first session of swimm

Romashka-Z-Leto [24]

24 girls

60% is 0.6
40 x 0.6 = 24

8 0

3 years ago

Read 2 more answers

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

6 inches how many centimeters. (Round to the nearest tenth if needed)

Artist 52 [7]

15.24 cm rounded to 15.2 cm

8 0

3 years ago

Read 2 more answers

6. Write the inequality using interval notation for the graph.

makkiz [27]

Answer:

Step-by-step explanation:

C is the correct answer because it represents exactly what the graph shows with the bold lines.

As you can see in the graph, there is no interception between 7 and 12. Therfore, the answer can't be B(because it represents the area between the two numbers).

A isn't the correct answer either because it represents a dot, which is obviously not the case here since there is no dot.

I can't explain why C isn't the right answer, but I hope that explanation that I gave is enough for you to understand ;)

(If you have any questions about the way I solved it, don't hesitate to ask me in the comments, and if you find my answer helpful - consider marking it Brainliest ÷) )

6 0

3 years ago

Brainliest goes to whoever answers correctly

lord [1]

The upper left and bottom right are functions.

Remember a function has to pass the vertical line test. Elements from the domain cannot repeat, but elements from the range can.

4 0

3 years ago