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2 years ago

PLEASE HELP!! I worked it out and haven’t found the right answer

Mathematics

1 answer:

8_murik_8 [283]2 years ago

5 0

Answer:

option B is correct. Once have a look to this solution that I have answered

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Find the point P that is 2/5 of the way from A to B on the directed line segment AB

Kruka [31]

The point P(–4, 4) that is $\frac{2}{5}$ of the way from A to B on the directed line segment AB.

Solution:

The points of the line segment are A(–8, –2) and B(6, 19).

P is the point that bisect the line segment in $\frac{2}{5}$ .

So, m = 2 and n = 5.

$x_1=-8, y_1=-2, x_2=6, y_2=19$

By section formula:

$$P(x, y)=\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)$

$$P(x, y)=\left(\frac{2\times 6+5\times (-8)}{2+5}, \frac{2\times 19+5\times (-2)}{2+5}\right)$

$$P(x, y)=\left(\frac{12-40}{7}, \frac{38-10}{7}\right)$

$$P(x, y)=\left(\frac{-28}{7}, \frac{28}{7}\right)$

P(x, y) = (–4, 4)

Hence the point P(–4, 4) that is $\frac{2}{5}$ of the way from A to B on the directed line segment AB.

4 0

3 years ago

Convert 3.5km to meters

tatuchka [14]

The answer is 3500 meters.
1 km = 1000 meters. This means that 3.5 x 1000 = 3500

6 0

2 years ago

How do you do proofs

AnnyKZ [126]

Answer:

you can watch YT videos or search online. search: How to solve proofs. Also search up proof theorems. This will help you even more towards solving more complex proof problems.

Step-by-step explanation:

8 0

2 years ago

World war 1 lasted from August 4th 1914 until November 11th, 1918. During this time an estimated 10 million lives were lost. Abo

Kryger [21]

The question is so dry, mechanical, and devoid of emotion
that it's terrifying.

There is no way to assign a number to "How many people were
dying per day", and I would prefer not even to think about it in
those terms.

-- The period of time from August 4, 1914 until November 11, 1918 is 1,560 days.

-- The "average", or better, the "unit rate" of 10 million events in 1,560 days
is the quotient

   (10,000,000 events) / (1,560 days)

   = 6,410.3 events per day

   = 267.1 events per hour

   =   4.45 events per minute.

Reciprocally, this is a unit rate of

13.48 seconds per event,
      sustained continuously for 4.274 years !

When will we ever learn ! ?

6 0

2 years ago

Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where

z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704± $2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}}$ ≈ -0.171 ±0.237 that is (−0.408, 0.066)

7 0

2 years ago