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larisa86 [58]
3 years ago
7

Please help me Find product

Mathematics
2 answers:
MrMuchimi3 years ago
5 0

Answer:

6t^2-5tsu+s^2u^2

Step-by-step explanation:

liraira [26]3 years ago
4 0

Answer:

s^2u^2−5stu+6t^2

Step-by-step explanation:

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Which step is included in the graph of the function f(x)=[x-1]? (the brackets are ceiling functions symbols)
KiRa [710]

we are given

f(x)=[x=1]

where bracket means ceiling functions

we know that

Ceiling  function returns the least value of the integer that is greater than or equal to the specified number

so, we can check each options

option-A:

-4\leq x

At x=-4:

f(x)=[-4-1] =-5

For x<-3:

Let's assume

x=-3.1

f(x)=[-3.1-1] =[-4.1]=-5

so, this interval is TRUE

option-B:

-2\leq x

At x=-2:

f(x)=[-2-1] =-3

For x<-1:

Let's assume

x=-1.1

f(x)=[-1.1-1] =[-2.1]=-3

so, this is FALSE

7 0
4 years ago
Read 2 more answers
What is the range of f(x)?
slavikrds [6]
It -2,-1 I hope this help
6 0
3 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
One number is 5 times another. If their <br><br>difference is 96, find the numbers.
san4es73 [151]
120 and 24 are the numbers
4 0
3 years ago
Can someone help me on this pls?
Svetllana [295]

Answer:

2 2/5

Step-by-step explanation:

7 0
3 years ago
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