Tommorow I have maths exam can anyone help me please grade 9 cbse

Consider an espresso stand with a single barista. Customers arrive at the stand at the rate of 28 per hour according to a Poisso

Snowcat [4.5K]

Answer:

The average number of customers in the system is 3.2

Step-by-step explanation:

The average number of customes in the system is given by:

$L = \frac{\lambda^{2}}{\mu(\mu - \lambda)}$

In which

$\lambda$ is the number of arirvals per time period

$\mu$ is the average number of people being served per period.

The number of arrivals is modeled by the Poisson distribution, while the service time is modeled by the exponential distribution.

Customers arrive at the stand at the rate of 28 per hour

This means that $\lambda = 28$

Service times are exponentially distributed with a service rate of 35 customers per hour.

This means that $\mu = 35$ . So

The average number of customers in the system (i.e., waiting and being served) is

$L = \frac{\lambda^{2}}{\mu(\mu - \lambda)}$

$L = \frac{28^{2}}{35(35 - 28)} = 3.2$

The average number of customers in the system is 3.2

4 0

3 years ago

normally distributed with mean 24.6 seconds and standard deviation 0.64 seconds;the lower finishing time, the better. If the qua

marshall27 [118]

Answer:

$z=1.04$

And if we solve for a we got

$a=24.6 +1.04*0.64=25.27$

So the value of height that separates the bottom 85% of data from the top 15% is 25.27.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the finishing time of a population, and for this case we know the distribution for X is given by:

$X \sim N(24.6,0.64)$

Where $\mu=24.6$ and $\sigma=0.64$

Solution to the problem

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.15$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.04. On this case P(Z<1.04)=0.85 and P(z>1.04)=0.15

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.04$

And if we solve for a we got

$a=24.6 +1.04*0.64=25.27$

So the value of height that separates the bottom 85% of data from the top 15% is 25.27.

3 0

3 years ago

What is the answer to x+4=4x-17

Lesechka [4]

Answer:

x=7

Step-by-step explanation:

x+4=4x-17

4=3x-17

21=3x

x=7

5 0

3 years ago

The U.S. Bureau of Economic Statistics reports that the average annual salary in the metropolitan Boston area is $50,542. Suppos

xenn [34]

Answer:

(a) P(X > $57,000) = 0.0643

(b) P(X < $46,000) = 0.1423

(c) P(X > $40,000) = 0.0066

(d) P($45,000 < X < $54,000) = 0.6959

Step-by-step explanation:

We are given that U.S. Bureau of Economic Statistics reports that the average annual salary in the metropolitan Boston area is $50,542.

Suppose annual salaries in the metropolitan Boston area are normally distributed with a standard deviation of $4,246.

Let X = annual salaries in the metropolitan Boston area

SO, X ~ Normal( $\mu=$50,542,\sigma^{2} = $4,246^{2}$ )

The z-score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma }$ ~ N(0,1)

where, $\mu$ = average annual salary in the Boston area = $50,542

$\sigma$ = standard deviation = $4,246

(a) Probability that the worker’s annual salary is more than $57,000 is given by = P(X > $57,000)

P(X > $57,000) = P( $\frac{X-\mu}{\sigma }$ > $\frac{57,000-50,542}{4,246 }$ ) = P(Z > 1.52) = 1 - P(Z $\leq$ 1.52)

= 1 - 0.93574 = 0.0643

The above probability is calculated by looking at the value of x = 1.52 in the z table which gave an area of 0.93574.

(b) Probability that the worker’s annual salary is less than $46,000 is given by = P(X < $46,000)

P(X < $46,000) = P( $\frac{X-\mu}{\sigma }$ < $\frac{46,000-50,542}{4,246 }$ ) = P(Z < -1.07) = 1 - P(Z $\leq$ 1.07)

= 1 - 0.85769 = 0.1423

The above probability is calculated by looking at the value of x = 1.07 in the z table which gave an area of 0.85769.

(c) Probability that the worker’s annual salary is more than $40,000 is given by = P(X > $40,000)

P(X > $40,000) = P( $\frac{X-\mu}{\sigma }$ > $\frac{40,000-50,542}{4,246 }$ ) = P(Z > -2.48) = P(Z < 2.48)

= 1 - 0.99343 = 0.0066

The above probability is calculated by looking at the value of x = 2.48 in the z table which gave an area of 0.99343.

(d) Probability that the worker’s annual salary is between $45,000 and $54,000 is given by = P($45,000 < X < $54,000)

P($45,000 < X < $54,000) = P(X < $54,000) - P(X $\leq$ $45,000)

P(X < $54,000) = P( $\frac{X-\mu}{\sigma }$ < $\frac{54,000-50,542}{4,246 }$ ) = P(Z < 0.81) = 0.79103

P(X $\leq$ $45,000) = P( $\frac{X-\mu}{\sigma }$ $\leq$ $\frac{45,000-50,542}{4,246 }$ ) = P(Z $\leq$ -1.31) = 1 - P(Z < 1.31)

= 1 - 0.90490 = 0.0951

The above probability is calculated by looking at the value of x = 0.81 and x = 1.31 in the z table which gave an area of 0.79103 and 0.9049 respectively.

Therefore, P($45,000 < X < $54,000) = 0.79103 - 0.0951 = 0.6959

3 0

3 years ago

3/7 to its decimal form and rounded to the nearest thousandth

ELEN [110]

3/7 is 0.42857142857 in decimal form. Rounded to the thousands its 0.429

5 0

3 years ago

Read 2 more answers