All categories

2 years ago

What is the slope of the line that passes through the points (−10,−8) and (-8, -16) Write your answer in simplest form.

Mathematics

1 answer:

jarptica [38.1K]2 years ago

7 0

Step-by-step explanation:

given,

point = ( -10, -8 ) and ( -8, -16 )

we know,

slope = y2 - y1 / x2 - x1

x1 , y1 = (-10 , -8 )

x2 , y2 = ( -8, -16 )

after inserting the values we got,

→ -16 - (-8 ) / -8 - (-10)

→ -8/2 → -4

therefore, slope of line is -4.

hope this answer helps you dear...take care and may this answer helps you dear!.

Top answer · 0 votes

f(g(x))=362−x−2k f(g(x))=x→362−x−2k=x→x2−2x(1−k)−4(k−9)=0 Solution for x are equal if and only if discriminant Δ=0 4(1−k)2+16(k−9)=0 ... More

3 0

3 years ago

Find the equation of the line perpendicular to y=3/2x-9 and passes through the point (6,2)

lys-0071 [83]

Answer:

y=-2/3x-2

Step-by-step explanation:

slope= -2/3

y=-2/3x+b

enter the given points

2=4+b

b=-2

y=-2/3x-2

6 0

2 years ago

Can someone help me out please

lbvjy [14]

16 is the answer jxhsjcjaks

5 0

2 years ago

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average re

Zolol [24]

Answer:

a) $df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

b) $df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

c) $df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X$ represent the sample mean

$s$ represent the standard deviation for the sample

$n$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:

Null hypothesis: $\mu \leq 20$

Alternative hypothesis: $\mu > 20$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

3 0

3 years ago