What's the square root of 2 rounded to the nearest tenth ?

Can someone help and I need it quick

kvasek [131]

Answer:

The change in momentum between 0 ms and 14 ms is 0 kg•m/s

Step-by-step explanation:

Here, we want to calculate the change in momentum using the graphical illustration

This is same as the slope using the two points

Let us have 2 expressions of (x,y)

Mathematically, that will be obtained from the graph as follows;

At 0 ms, we have 0 N

At 14 ms , we also have 0

So the two points we are considering are;

(0,0) and (14,0)

= (0-0)/(14-0) = 0/14 = 0 N/ms

8 0

3 years ago

What does it mean by find the slope of each line

beks73 [17]

Answer:

.

Step-by-step explanation

Find the run and the rise.

To do that:

- There is a right angle where the points meet. The bottom one (2,-2) is three units away from the top one, (-2,-1)and then one upwards and they will meet.

6 0

3 years ago

Find the mean and median for this set of number: 80, 70, 60, 40, 80.

castortr0y [4]

Median: 70 mean: 55.

5 0

3 years ago

Read 2 more answers

A researcher was interested in comparing the resting heart rate of people who exercise regularly and people who do not exercise

alekssr [168]

Answer:

$t=\frac{(73.5-69.3)-0}{\sqrt{\frac{10.2^2}{16}+\frac{8.7^2}{12}}}}=1.173$

Using the critical value method we need to find a critical value in the t distribution who accumulates 0.025 of the area on the right and we got:

$t_{cric}= 2.06$

So then since our calculated value is lower then the critical value we fail to reject the null hypothesis and we can't conclude that the true mean resting pulse rate of people who do not exercise regularly is larger than the mean resting pulse rate of people who exercise regularly.

Step-by-step explanation:

Information given

$\bar X_{1}=73.5$ represent the mean for sample of people who do not exercise

$\bar X_{2}=69.3$ represent the mean for sample of people who do exercise

$s_{1}=10.2$ represent the sample standard deviation for 1

$s_{2}=8.7$ represent the sample standard deviation for 2

$n_{1}=16$ sample size for the group 1

$n_{2}=12$ sample size for the group 2

$\alpha=0.01$ Significance level

t would represent the statistic

System of hypothesis

We want to verify if the mean resting pulse rate of people who do not exercise regularly is larger than the mean resting pulse rate of people who exercise regularly, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2} \leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

The statistic for this case would be given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

The degrees of freedom are given by:

$df=n_1 +n_2 -2=16+12-2=26$

Replacing the info given we got:

$t=\frac{(73.5-69.3)-0}{\sqrt{\frac{10.2^2}{16}+\frac{8.7^2}{12}}}}=1.173$

Using the critical value method we need to find a critical value in the t distribution who accumulates 0.025 of the area on the right and we got:

$t_{cric}= 2.06$

So then since our calculated value is lower then the critical value we fail to reject the null hypothesis and we can't conclude that the true mean resting pulse rate of people who do not exercise regularly is larger than the mean resting pulse rate of people who exercise regularly.

6 0

3 years ago

Use the quadratic formula x^2-5x+3=0

kifflom [539]

Answer:

$x^2 -5x +3 = 0\\\\ x= \frac{-b \pm \sqrt{b^2-4*a*c} }{2a} \\\\b = -5, a = 1 , c = 3\\\\x= \frac{-b + \sqrt{b^2-4*a*c} }{2a} , x= \frac{-b -\sqrt{b^2-4*a*c} }{2a}\\\\x= \frac{5 + \sqrt{25-12} }{2}, x= \frac{5-\sqrt{25-12} }{2}\\\\x= \frac{5+\sqrt{13} }{2}, x= \frac{5- \sqrt{13} }{2}$

5 0

2 years ago