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3 years ago

Probability of rolling a factor of 45

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

7 0

Answer:

Step-by-step explanation:

POOP THEN PEEEEEE

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3x-7y=17 when x=-3,0,3

NeX [460]

$3x-7y=17\\-7y=17-3x\ \ \ \ |:(-7)\\y=-\frac{17}{7}+\frac{3}{7}x\\\\When\ x=-3:\\y=-\frac{17}{7}+\frac{3}{7}\cdot(-3)\\y=-\frac{17}{7}-\frac{9}{7}\\y=-\frac{26}{7}=-3\frac{5}{7}\\\\When\ x=0:\\y=-\frac{17}{7}+\frac{3}{7}\cdot0\\y=-\frac{17}{7}=-2\frac{3}{7}\\\\When\ x=3:\\y=-\frac{17}{7}+\frac{3}{7}\cdot3\\y=-\frac{17}{7}+\frac{9}{7}\\y=-\frac{8}{7}=-1\frac{1}{7}$

4 0

3 years ago

A random sample of 4"4 fields" of barley has a mean yield of "49.6"49.6 bushels per acre and standard deviation of 7.997.99 bush

ExtremeBDS [4]

Answer:

The critical value that should be used in constructing the interval is T = 5.8408.

The 99% confidence interval for the true mean yield is between 2.943 bushels per acre and 96.268 bushels per acre.

Step-by-step explanation:

We have the standard deviation of the sample, so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 4 - 1 = 3

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 3 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 5.8408. This is the critical value.

The margin of error is:

M = T*s = 5.8408*7.99 = 46.668 bushels per acre

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 49.6 - 46.668 = 2.943 bushels per acre.

The upper end of the interval is the sample mean added to M. So it is 49.6 + 46.668 = 96.268 bushels per acre.

The 99% confidence interval for the true mean yield is between 2.943 bushels per acre and 96.268 bushels per acre.

8 0

4 years ago

mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi

chubhunter [2.5K]

Hello!

We have the following data:

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

$a_n = a_1 + (n-1)*r$

* second year salary

$a_2 = a_1 + (2-1)*300$

$a_2 = 32000 + 1*300$

$a_2 = 32000 + 300$

$\boxed{a_2 = 32300}$

* third year salary

$a_3 = a_1 + (3-1)*300$

$a_3 = 32000 + 2*300$

$a_3 = 32000 + 600$

$\boxed{a_3 = 32600}$

* fourth year salary

$a_4 = a_1 + (4-1)*300$

$a_4 = 32000 + 3*300$

$a_4 = 32000 + 900$

$\boxed{a_4 = 32900}$

* fifth year salary

$a_5 = a_1 + (5-1)*300$

$a_5 = 32000 + 4*300$

$a_5 = 32000 + 1200$

$\boxed{a_5 = 33200}$

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

$a_6 = a_1 + (6-1)*300$

$a_6 = 32000 + 5*300$

$a_6 = 32000 + 1500$

$\boxed{a_6 = 33500}$

* seventh year salary

$a_7 = a_1 + (7-1)*300$

$a_7 = 32000 + 6*300$

$a_7 = 32000 + 1800$

$\boxed{a_7 = 33800}$

* eighth year salary

$a_8 = a_1 + (8-1)*300$

$a_8 = 32000 + 7*300$

$a_8 = 32000 + 2100$

$\boxed{a_8 = 34100}$

* ninth year salary

$a_9 = a_1 + (9-1)*300$

$a_9 = 32000 + 8*300$

$a_9 = 32000 + 2400$

$\boxed{a_9 = 34400}$

* tenth year salary

$a_{10} = a_1 + (10-1)*300$

$a_{10} = 32000 + 9*300$

$a_{10} = 32000 + 2700$

$\boxed{a_{10} = 34700}$

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

* eleventh year salary

$a_{11} = a_1 + (11-1)*300$