What kind of sequence is this? 52, 39, 26, 13, ...

Write the standard form of the line that passes through the point (-2, 4) and is parallel to x - 2y = 6.

Radda [10]

First calculate the function of x-2y=6:
x-2y=6 | - x
-2y=6-x | ÷ (-2)
y=0.5x-3

Now you have the function. Because the slope you search is parallel to this one, the slope is the same (in this case 0.5). A slope of 0.5 means for 2 steps to the right go one up. To define the function you are searching you have to find out the y intercept. For this take your point (-2|4). Now go as many steps to the right until you reach the y intercept or x=0 in this case you need 2 steps. So for 2 steps right you go one up. Now you have the y intercept (5).
The resulting function is: f(x) = 0.5+6.

6 0

2 years ago

30 points! PLS HELP IF U SMART

ANTONII [103]

Answer:

y = 40 x = 80

Step-by-step explanation:

520 = 3.25x + 6.5y and x+y=120

x+y=120

x = 120-y

520 = 3.25(120-y)+6.5y

520 = 390-3.25y+6.5y

520 = 390+3.25y

130=3.25y

40 = y

x = 120-y

x = 120-(40)

x = 80

here's a graph

8 0

2 years ago

Read 2 more answers

Show all possible arrangements of the letters in the word FRED using a tree diagram. If each of the letters is on a tile and dra

Tanzania [10]

Answer:

24

Step-by-step explanation:

F R E D

F R D E

F E R D

F E D R

F D E R

F D R E

so for each letter u have 6 so 4 x 6=24

4 0

2 years ago

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng

alexandr402 [8]

Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.

b. A 95% confidence interval for $\mu$ .

Answer:

a

$\= x = 13.4$ . $E = 2.3$

b

$10.7 < \mu < 16.1$

Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The 90% confidence interval is (11.1, 15.7)

Generally the point estimate of $\mu$ is mathematically evaluated as

$\= x = \frac{11.1 + 15.7 }{2}$

=> $\= x = 13.4$

Generally the margin of error is mathematically evaluated as

$E = \frac{15.7 - 11.1}{2 }$

=> $E = 2.3$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the equation for the lower limit of the confidence interval is

$\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1$

=> $13.4 - 0.3877 s = 11.1$

=> $s = 5.932$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * \frac{5.932}{\sqrt{18} }$

=> $E = 2.7$

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $13.4 - 2.7 < \mu < 13.4 + 2.7$

=> $10.7 < \mu < 16.1$

3 0

2 years ago

According to a 2014 research study of national student engagement in the U.S., the average college student spends 17 hours per w

Nadusha1986 [10]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 17

For the alternative hypothesis,

µ < 17

This is a left tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 80,

Degrees of freedom, df = n - 1 = 80 - 1 = 79

t = (x - µ)/(s/√n)

Where

x = sample mean = 15.6

µ = population mean = 17

s = samples standard deviation = 4.5

t = (15.6 - 17)/(4.5/√80) = - 2.78

We would determine the p value using the t test calculator. It becomes

p = 0.0034

Since alpha, 0.05 > than the p value, 0.0043, then we would reject the null hypothesis.

The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.

4 0

3 years ago