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3 years ago

HELP PLEASE FAST !!!

Mathematics

2 answers:

EastWind [94]3 years ago

6 0

Answer:

0.92 cents

Step-by-step explanation:

Alexandra [31]3 years ago

4 0

Answer:

$37.97 - $37.05 = 0.92 cents

Step-by-step explanation:

You might be interested in

Assume that SAT scores are normally distributed with mean 1518 and standard deviation 325. Round your answers to 4 decimal place

Katyanochek1 [597]

Answer:

a. 0.2898

b. 0.0218

c. 0.1210

d. 0.1515

e. This is because the population is normally distributed.

Step-by-step explanation:

Assume that SAT scores are normally distributed with mean 1518 and standard deviation 325. Round your answers to 4 decimal places

We are using the z score formula when random samples

This is given as:

z = (x-μ)/σ/√n

where x is the raw score

μ is the population mean

σ is the population standard deviation.

n is the random number of samples

a.If 100 SAT scores are randomly selected, find the probability that they have a mean less than 1500.

For x = 1500, n = 100

z = 1500 - 1518/325/√100

z = -18/325/10

z = -18/32.5

z = -0.55385

Probability value from Z-Table:

P(x<1500) = 0.28984

Approximately = 0.2898

b. If 64 SAT scores are randomly selected, find the probability that they have a mean greater than 1600

For x = 1600, n = 64

= z = 1600 - 1518/325/√64.

z= 1600 - 1518 /325/8

z = 2.01846

Probability value from Z-Table:

P(x<1600) = 0.97823

P(x>1600) = 1 - P(x<1600) = 0.021772

Approximately = 0.0218

c. If 25 SAT scores are randomly selected, find the probability that they have a mean between 1550 and 1575

For x = 1550, n = 25

z = 1550 - 1518/325/√25

z = 1550 - 1518/325/5

z = 1550 - 1518/65

= 0.49231

Probability value from Z-Table:

P(x = 1550) = 0.68875

For x = 1575 , n = 25

z = 1575 - 1518/325/√25

z = 1575 - 1518/325/5

z = 1575 - 1518/65

z = 0.87692

Probability value from Z-Table:

P(x=1575) = 0.80974

The probability that they have a mean between 1550 and 1575

P(x = 1575) - P(x = 1550)

= 0.80974 - 0.68875

= 0.12099

Approximately = 0.1210

d. If 16 SAT scores are randomly selected, find the probability that they have a mean between 1440 and 1480

For x = 1440, n = 16

z = 1440 - 1518/325/√16

= -0.96

Probability value from Z-Table:

P(x = 1440) = 0.16853

For x = 1480, n = 16

z = 1480 - 1518/325/√16

=-0.46769

Probability value from Z-Table:

P(x = 1480) = 0.32

The probability that they have a mean between 1440 and 1480

P(x = 1480) - P(x = 1440)

= 0.32 - 0.16853

= 0.15147

Approximately = 0.1515

e. In part c and part d, why can the central limit theorem be used even though the sample size does not exceed 30?

The central theorem can be used even though the sample size does not exceed 30 because the population is normally distributed.

6 0

3 years ago

Write an equation in slope intercept form for the graph of the line.

olganol [36]

Answer:

y = 2x - 3

Step-by-step explanation:

Slope intercept form:

y = mx + b, where m - slope, b - y-intercept

Take two points from the graph:

(2, 1) and (3, 3)

Find the slope:

m = (3 - 1)/(3 - 1) = 2/1 = 2

Find the y-intercept using point (2, 1) and the equation above:

1 = 2*2 + b
b = 1 - 4
b = -3

The line is:

y = 2x - 3

5 0

2 years ago

Read 2 more answers

Simplify the following:

marin [14]

Answer:

no one can do it

Step-by-step explanation:

6 0

2 years ago

Rectangle has a vertices at (-1,4), (-1,-1) and (4, -1) where is the fourth vertex located

erica [24]

Answer:

(4,4)

Step-by-step explanation:

1. I drew a grid and plotted the 3 points I already knew

2. I found the spot that had the same x coordinate that the point to the bottom left of the shape, and the same y coordinate of the top right of the shape which was the point (4,4)

I hope this helps!!

3 0

3 years ago

An insurance company sells automobile liability and collision insurance. Let X denote the percentage of liability policies that

qaws [65]

Answer:

-764.28

Step-by-step explanation:

Given the joint cumulative distribution of X and Y as

$F(x,y) = \frac{xy(x+y)}{2000000}\ \ \ \, \ 0\leq x100, 0\leq y\leq 100$

#First find $F_x$ and probability distribution function , $f_x(x)$ :

$F_x(x)=F(x,100)\\\\\\=\frac{100x(x+100)}{2000000}\\\\\\\\=\frac{100x^2+10000x}{2000000}\\\\\\=>f_x(x)=\frac{x}{10000}+\frac{1}{200}$

#Have determined the probability distribution unction , $f_x(x)$ , we calculate the Expectation of the random variable X:

$E(X)=\int\limits^{100}_0 \frac{x^2}{10000}+\frac{x}{200} dx \\\\\\\\=|\frac{x^3}{30000}+\frac{x^2}{400}|\limits^{100}_0\\\\=58.33\\\\$

#We then calculate $E(X^2)$ :

$E(X^2)=\int\limits^{100}_0 \frac{x^3}{10000}+\frac{x^2}{200}\ dx\\\\=\frac{x^4}{40000}+\frac{x^3}{600}|\limits^{100}_0=4166.67\\\\Var(X)=E(X^2)-(E(X))^2=4166.67-58.33^2\\\\Var(X)=764.28$

Hence, the Var(X) is 764.28

4 0

3 years ago