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Aleksandr-060686 [28]

2 years ago

11

Point J(-2, 1) and point K(4, 5) form the line segment jk. for the point p that partitions jk in the ratio 3:7 what is the y coo

rdinate of p

1 answer:

kumpel [21]2 years ago

7 0

Answer:

$\frac{11}{5}$

Step-by-step explanation:

Using the section formula

$y_{P}$ = $\frac{3(5)+7(1)}{3+7}$ = $\frac{15+7}{10}$ = $\frac{22}{10}$ = $\frac{11}{5}$

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-. For the event of rolling three standard fair 6-sided dice, find the following probabilities. Give all

elixir [45]

Answer:

1/6 or 6/36

Step-by-step explanation:

Given the qualifications, there are six qualifying combinations out of the 36 combinations, which simplified, is 1 out of 6.

4 0

2 years ago

What is the greatest common factor for 30,48,60 and whats the GCF ??????

Svetach [21]

30 = 2 * 3 * 5

<span>48 = 2⁴ * 3
</span>
<span>60 = 2² * 3 * 5

</span><span>GCF = 2 * 3 = 6</span>

4 0

3 years ago

Read 2 more answers

1.15 GPA and study hours: A survey was conducted on 193 Duke University undergraduates who took an introductory statistics cours

KonstantinChe [14]

Answer:

Step-by-step explanation:

A

According to the exercise we can say that the explanatory variable will be hours / weeks of study.

The response variable is GPA.

B

We definitely found that between the variables there is a weak or fragile relationship, neither negative nor positive nor linear.

C)

1.Observational study, since we observe that the only variables that will influence the results of the study are the observations of the students

2.No. This study is observational and its characteristics are not sufficient to conclude that studying more hours leads to obtaining a higher GPA.

8 0

3 years ago

The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m st

Reil [10]

Answer:

a) 30.726m/s and b) 5.5549s

Step-by-step explanation:

a.) What was Chris Huber’s speed in meters per second(m/s)?

Given the distance and time, the formula to obtain the speed is

$v=\frac{d}{t}$ .

Applying this to our problem we have that

$v=\frac{200m}{6.509s}= 30.726m/s$ .

So, Chris Huber’s speed in meters per second(m/s) was 30.726m/s.

b) What was Whittingham’s time through the 200 m?

In a) we stated that $v=\frac{d}{t}$ . This formula implies that

$t=\frac{d}{v}$ .

First, observer that $19\frac{km}{h} =19,000\frac{m}{h}=\frac{19,000}{3,600}m/s= 5.2777m/s$ .

Then, Sam Whittingham speed was equal to Chris Huber’s speed plus 5.2777 m/s. So, $v=30.726\frac{m}{s} +5.2777\frac{m}{s}= 36.003 m/s.$

Then, applying 1) we have that

$t=\frac{200m}{36.003m/s}=5.5549s.$

So, Sam Whittingham’s time through the 200 m was 5.5549s.

5 0

3 years ago

Express 0.001 in scientific notation by counting decimal places

kati45 [8]

1 x 10-3 make sure the -3 is above the 10.

7 0

3 years ago