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2 years ago

5

Holly bought a magazine subscription for a year. She paid $27. Holly wanted to find the price, p, of the subscription each month

. She created the model shown to help find this price. 27 р |р |р |р |р |р |р |р | Р | Р |р | p What was the price of the subscription each month? F $39.00 G $2.25 H $324.00 3 $22.50

1 answer:

sweet-ann [11.9K]2 years ago

3 0

Answer:

2.25 should be the right answer

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The quotient of n and 12 subtracted from 100?

Korvikt [17]

Answer:

100 - 12/n

Step-by-step explanation:

quotient means division so

12/n

and subtracted from means say x - 12/n so

100 - 12/n

If you're looking to evaluate it then it would be

4(25n - 3) / n

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2 years ago

marin [14]

Answer:

4 feet of wall should be on each side of the painting.

Step-by-step explanation:

Given that a 4 foot wide painting should be centered on a 12 foot wide wall, to determine how many feet (X) should be on each side of the painting, the following equation must be solved:

(Total wide of the wall - wide of the painting) / 2 = X

(12 - 4) / 2 = X

8/2 = X

4 = X

Therefore, 4 feet of wall should be on each side of the painting.

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3 years ago

I only need 12 and 17, first good response i will give brainliest please i’m desperate

Mumz [18]

Answer:

$\textsf{12.} \quad y = 6x - 5$

$\textsf{17.} \quad y=-\dfrac{1}{4}x-2$

Step-by-step explanation:

<h3><u>Question 12</u></h3>

Find the slope of the line by substituting two points from the given table into the slope formula.

<u>Define the points</u>:

Let (x₁, y₁) = (2, 7)
Let (x₂, y₂) = (3, 13)

$\implies \textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{13-7}{3-2}=\dfrac{6}{1}=6$

Substitute the found slope and point (2, 7) into the point-slope formula to create an equation of the line:

$\implies y-y_1=m(x-x_1)$

$\implies y-7=6(x-2)$

$\implies y-7=6x-12$

$\implies y=6x-5$

<h3><u>Question 17</u></h3>

Given:

f(4) = 3
f(0) = -2

Therefore, two points on the line are:

(4, -3)
(0, -2)

The y-intercept is the y-value when x = 0.

Therefore, the y-intercept of the line is -2.

$\boxed{\begin{minipage}{6.3 cm}\underline{Slope-intercept form of a linear equation}\\\\$y=mx+b$\\\\where:\\ \phantom{ww}$\bullet$ $m$ is the slope. \\ \phantom{ww}$\bullet$ $b$ is the $y$-intercept.\\\end{minipage}}$

Substitute the y-intercept and the point (4, 3) into the slope-intercept formula and solve for <em>m</em> to find the slope:

$\implies y=mx+b$

$\implies -3=m(4)-2$

$\implies -1=4m$

$\implies m=-\dfrac{1}{4}$

Therefore, the equation of the line is:

$y=-\dfrac{1}{4}x-2$

7 0

11 months ago

Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

<em>Let </em> $\mu_1$ <em> = true mean number of sales at location A.</em>

<em /> $\mu_2$ = <em>true mean number of sales at location B</em>

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, <u><em>test statistics</em></u> = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

<em>Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

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3 years ago

10 times as much as 2,000 is 1/10 of

Natali5045456 [20]

20,000 because 20,000 is ten times as much as 2,000 and 2,000 is 1/10 of 20,000

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2 years ago