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amm1812
3 years ago
9

Find the slope between these two points: (3, -10) and (-2, -30). Show all of your work. what the answer

Mathematics
2 answers:
kari74 [83]3 years ago
6 0

Answer:

4

Step-by-step explanation:

slope equation is y2-y1/x2-x1

insert points: -30-(-10)/-2-3

simplify: -30+10/-5

simplify further: -20/-5

negatives cancel because there cannot be both on top, making it positive: 20/5

divide: 20/5=4

Harlamova29_29 [7]3 years ago
3 0

\huge\text{$m=\boxed{4}$}

Hey there! Start with the slope formula, where (x_1,y_1) and (x_2,y_2) are the two known points.

\begin{aligned}m&=\dfrac{y_2-y_1}{x_2-x_1}\\&=\frac{-30-(-10)}{-2-3}\end{aligned}

Simplify.

\begin{array}{c|l}\textbf{Solving}&\textbf{Reason}\\\cline{1-2}\\m=\dfrac{-30+10}{-2-3}&x-(-y)=x+y\\\\m=\dfrac{-20}{-5}&\text{Addition and subtraction}\\\\m=\dfrac{20}{5}&\text{The negatives cancel out}\\\\m=\dfrac{4}{1}&\text{Divide the numerator and denominator by $5$}\\\\m=\boxed{4}&\dfrac{x}{1}=x\end{aligned}

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Answer:

26 - \sqrt{181} cm

Step-by-step explanation:

The volume of the box is:

V = height * length * width

V = x*(66 - 2*x)*(90 - 2*x)

V = (66*x - 2*x^2)*(90 - 2*x)

V = 5940*x - 132*x^2 - 180*x^2 + 4*x^3

V = 4*x^3 - 312*x^2 + 5940*x

where x is the length of the sides of the squares,  in cm.

The mathematical problem is :

Maximize: V = 4*x^3 - 312*x^2 + 5940*x

subject to:

x > 0

2*x < 66 <=> x < 33

In the maximum, the first derivative of V, dV/dx, is equal to zero

dV/dx = 12*x^2 - 624*x + 5940

From quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}

x = \frac{624 \pm \sqrt{(-624)^2 - 4(12)(5940)}}{2(12)}

x = \frac{624 \pm \sqrt{104256}}{24}

x = \frac{624 \pm \sqrt{2^6*3^2*181}}{24}

x = \frac{624 \pm 8*3*\sqrt{181}}{24}

x_1 = \frac{624 + 24*\sqrt{181}}{24}

x_1 = 26 + \sqrt{181}

x_2 = \frac{624 - 24*\sqrt{181}}{24}

x_2 = 26 - \sqrt{181}

But x_1 > 33, then is not the correct answer.

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Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

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Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

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The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

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Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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