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2 years ago

13

Find the midpoint (9,1), (8,-4)

1 answer:

Aleks04 [339]2 years ago

3 0

Answer:

Midpoint (9,1,(8-4) = (8.5, -1.5)

step-by step:Midpoint (9,1, 8-4)=(xA+xB2,yA+yB2)=(9+82,1+−42)=(172,−32)Midpoint of a line segment (9,1, 8-4)= (8.5, -1.5)

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with Judy jogging north and Jackie jogging west (on straight roads). When Jackie is 1 mile farther fro m the intersection than J

miskamm [114]

Let the distance of Judy from intersection is x and distance of Jackie from intersection is y.

We convert the given information to equations, step by step.

Point 1: When Jackie is 1 mile farther from the intersection than Judy.

This means y is 1 mile more than x.

So,

y = 1 + x

Point 2: The distance between them is 2 miles more than Judy’s distance from the intersection.

Distance between is x+ y.

So, x+y is 2 miles more than y.

x+y = y + 2
⇒
x = 2

From point 1 we have:

y = 1 + x = 1+ 2 = 3

So,

Distance of Judy from intersection is 2 miles and distance of Jackie from intersection is 3 miles.

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2 years ago

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2 years ago

Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust Jet red Cloudtran

$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

3 0

2 years ago

Calculate the average rate of change of f(x) = x2 4 - 5 for 3 ≤ x ≤ 5.

pogonyaev

Answer:

8

Step-by-step explanation:

Substitute the equation and values into the average rate of change formula.

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3 years ago

What is 7 divided by 1/3 in multiplication

sveta [45]

It would be 7x3 which would equal 21. Hope that helped!

4 0

2 years ago

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