Divide 32 students into 2 groups so the ratio is two to five how many students with be in each group

Write the equation of a line in slope-intercept form that has a slope of -0.5 and passes through the point (-5, 1.5)

baherus [9]

Y = -1/5x + 0.5
The y intercept is 0.5

3 0

2 years ago

Compute the standard deviation of the following set of data to the nearest whole number:

Klio2033 [76]

Answer:

Step-by-step explanation:

The standard deviation is the square root of the variance, and the variance is found by using the mean. So we will do that first. I will use the population variance as opposed to the sample variance since our set of numbers is small.

The mean: 8 + 12 + 15 + 17 + 18 = 70 and divide that by 5 to get

$\bar{x}=14$ and use this to find the variance in the formula:

$s^2=\frac{\sum(x_i-\bar{x})^2}{n}$ it is a bit difficult to enter that formula in correctly, but here's how it works mathematically:

$s^2=\frac{(8-14)^2+(12-14)^2+(15-14)^2+(17-14)^2+(18-14)^2}{5}$

Squaring this ensures us that we don't end up with zero, which would be useless.

$s^2=\frac{36+4+1+9+16}{5}$ so

$s^2=13.2$ which means that the standard deviation is

s = 3.633

(If you do it with n-1 = 4 in the denominator of the variance, you get a standard deviation of 4.062)

3 0

2 years ago

QUESTION 3 [10 MARKS] A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the nu

Harman [31]

Answer:

(a)Revenue function, $R(x)=580x-x^2$

Marginal Revenue function, R'(x)=580-2x

(b)Fixed cost =900 .

Marginal Cost Function=300+50x

(c)Profit, $P(x)=-35x^2+280x-900$

(d)x=4

Step-by-step explanation:

Part A

Price Function $= 580 - 10x$

The revenue function

$R(x)=x\cdot (580-10x)\\R(x)=580x-x^2$

The marginal revenue function

$\dfrac{dR}{dx}= \dfrac{d}{dx}(R(x))=\dfrac{d}{dx}(580x-x^2)=580-2x\\R'(x)=580-2x$

Part B

(Fixed Cost)

The total cost function of the company is given by $c=(30+5x)^2$

We expand the expression

$(30+5x)^2=(30+5x)(30+5x)=900+300x+25x^2$

Therefore, the fixed cost is 900 .

Marginal Cost Function

If $c=900+300x+25x^2$

Marginal Cost Function, $\frac{dc}{dx}= (900+300x+25x^2)'=300+50x$

Part C

Profit Function

Profit=Revenue -Total cost

$580x-10x^2-(900+300x+25x^2)\\580x-10x^2-900-300x-25x^2\\$Profit,P(x)=-35x^2+280x-900$

Part D

To maximize profit, we find the derivative of the profit function, equate it to zero and solve for x.

$P(x)=-35x^2+280x-900\\P'(x)=-70x+280\\-70x+280=0\\-70x=-280\\$Divide both sides by -70\\x=4$

The number of cakes that maximizes profit is 4.

6 0

2 years ago

Suppose a random sample of 50 college students are asked to measure the length of their right foot in centimeters. A 95% confide

hichkok12 [17]

Answer:

A 99% confidence interval will be wider than a 95% confidence interval

Step-by-step explanation:

From the question we are told that

The 95% confidence interval for for the mean foot length for students at the college is found to be 21.709 to 25.091 cm

Generally the width of a confidence interval is dependent on the margin of error.

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

From the above equation we see that

$E \ \ \alpha \ \ Z_{\frac{\alpha }{2} }$

Here $Z_{\frac{\alpha }{2} }$ is the critical value of the half of the level of significance and this value increase as the confidence level increase

Now if a 99% confidence level is used , it then means that the value of

$Z_{\frac{\alpha }{2} }$ will increase, this in turn will increase the margin of error and in turn this will increase the width of the confidence interval

Hence a 99% confidence interval will be wider than a 95% confidence interval

5 0

2 years ago

Hii can someone help with this pls :)

BlackZzzverrR [31]

Answer:

$R(x,y) = (0.8,-1.2)$