Can some one help me i have no idea how to do this

If it takes Leaping Louie 5 minutes to jump 3 blocks, how long will it take for him to jump 15 blocks?

Svetach [21]

25 minutes
3 blocks x 5 = 15 blocks

Therefore,
5 minutes x 5 = 25 minutes

6 0

3 years ago

Question part points submissions used solve the given differential equation by separation of variables. dx + e2xdy = 0

lina2011 [118]

Try this:
if given dx+e²ˣdy=0, then
$dy=- \frac{dx}{e^{-2x}} \ =\ \textgreater \ \ \int dy=-\int e^{-2x}dx; \ =\ \textgreater \ \ y= \frac{1}{2}e^{-2x}+C;$

4 0

3 years ago

Four students were discussing how to find the unit rate for a proportional relationship. Which method is valid?

BigorU [14]

Answer:

A

Step-by-step explanation:

Look at the graph of the relationship. Find the y-value of the point that corresponds to x = 1. That value is the unit rate

7 0

3 years ago

Read 2 more answers

Pip was thinking of a number. Pip doubles it and gets the answer of 50.8. What was the original number?

DerKrebs [107]

Answer:

25.4

Step-by-step explanation:

If double of the number 'x' is 50.8, (2x=50.8)

Then the original number must be half of that value, (x=50.8/2)

50.8/2 = 25.4

8 0

2 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

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

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

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

3 years ago