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2 years ago

32 + (7 − 2) ⋅ 4 − 6 over 3.

Mathematics

2 answers:

UkoKoshka [18]2 years ago

7 0

1. Answer: 37

2. Answer: 2

Oliga [24]2 years ago

3 0

15.3 the 3 repeats itself FOREVER

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THIS IS URGENT, IM ABOUT TO START CRYING!!!

tresset_1 [31]

PART A:

Finding the slope of the function f(x)

Choose any two pairs of coordinate from the table; (-1, -15) and (0, -10)

Let (-1, -15) be (x₁, y₁) and (0, -10) be (x₂, y₂)

Slope =

Slope of f(x) = 5

The function g(x) is given in the straight line equation form

Where, is the slope and is the y-intercept

Slope of g(x) = 2

-----------------------------------------------------------------------------------------------------------

g(x) = 2x + 8

Where, the slope (m) = 2 and the y-intercept (c) = 8

The y-intercept of g(x) is 8

for f(x), we can read the y-intercept when x = 0.

From the table, when x = 0, y = -10

The y-intercept of f(x) is -10

Function g(x) has higher y-intercept

3 0

2 years ago

What is the volume of a sphere with a radius of 5.9 in, rounded to the nearest tenth of a cubic inch?

Lynna [10]

$\textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=5.9 \end{cases}\implies V=\cfrac{4\pi (5.9)^3}{3}\implies V\approx 860.3~in^3$

7 0

1 year ago

Plz with steps .. it's very hard can anyone plz

liubo4ka [24]

Answer:

Step-by-step explanation:

$\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a} \qquad f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a} \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} }{\dfrac{1}{2\sqrt{x}} }\\\\$

$\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}} =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\$

7 0

2 years ago

Which number line shows the solution set for |y-9| <12

RUDIKE [14]

Answer:

Option c

$y\geq-3$ or $y \leq 21$

Step-by-step explanation:

The absolute value is a function that transforms any value x into a positive number.

Therefore, for the function $f(x) = |x|$ x> 0 for all real numbers.

Then the inequation:

$|y-9| \leq 12$ has two cases

$(y-9)$ if $y \geq 9$ (i)

$-(y-9)$ if $y < 9$ (ii)

We solve the case (i)

$y \leq 9 + 12\\y\leq21$

We solve the case (ii)

$-y +9 \leq 12\\-y \leq 12-9\\y \geq -3$

Then the solution is:

$y\geq-3$ or $y \leq 21$

8 0

2 years ago

A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou

valkas [14]

Answer:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

Step-by-step explanation:

Information given

$\bar X=1.8$ represent the sample mean for the growth

$s=0.5$ represent the sample standard deviation

$n=48$ sample size

$\mu_o =2$ represent the value that we want to compare

$\alpha=0.01$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 2cm per week, the system of hypothesis are :

Null hypothesis: $\mu \geq 2$

Alternative hypothesis: $\mu < 2$

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

6 0

3 years ago