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3 years ago

13

PLS HELP ME ! ! ! ! !

1 answer:

emmainna [20.7K]3 years ago

8 0

Answer:

D (0, 2)

General Formulas and Concepts:

<u>Algebra I</u>

The y-intercept is the y value when x = 0. Another way to reword that is when the graph crosses the y-axis.

Slope-Intercept Form: y = mx + b

m - slope
b - y-intercept

Step-by-step explanation:

<u>Step 1: Define</u>

y = 3/4x + 2

<u>Step 2: Break Function</u>

<em>Identify Parts</em>

Slope <em>m</em> = 3/4

y-intercept <em>b</em> = 2

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PLEASE HELP ITS DUE AT 12!!!

Black_prince [1.1K]

Answer:

Question 1: p=4 b=5 n=1.5

Question 2:Y=-3x+2

Step-by-step explanation:

i think i did this in arush

6 0

2 years ago

Read 2 more answers

A country is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 30

kondaur [170]

Answer:

We conclude that speed is greater than 30 miles per hour.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30 miles per hour

Sample mean, $\bar{x}$ = 35 miles per hour

Sample size, n = 15

Alpha, α = 0.01

Sample standard deviation, s = 4.7 miles per hour

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 30\text{ miles per hour}\\H_A: \mu > 30\text{ miles per hour}$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{35 - 30}{\frac{4.7}{\sqrt{15}} } = 4.120$

Now, $t_{critical} \text{ at 0.01 level of significance, 14 degree of freedom } = 2.624$

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis and conclude that speed is greater than 30 miles per hour.

We calculate the p-value.

P-value = 0.00052

Since p value is lower than the significance level, we reject the null hypothesis and accept the alternate hypothesis. We conclude that speed is greater than 30 miles per hour.

6 0

3 years ago

1.) Multiply. (x – 7)(2x – 7)

drek231 [11]

hope it will help u......

4 0

3 years ago

Read 2 more answers

Please help and explain answer

prohojiy [21]

ABCD is square so AB = BC = CD = AD
10 - x = x - 8 so x= 9 then AD will be 1yd, i guess im not sure bc i dunno the 2x-17 for :((

5 0

3 years ago

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago