Colleen rode her bicycle 9.5 miles in 0.8 hour. What was her average speed in miles per hour?

X2 +10x=39 , how do i solve or whats the answer

stealth61 [152]

Answer:

x = 3,-13 :)

Step-by-step explanation:

hope this helps plz brainiest

4 0

3 years ago

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed

Natali [406]

Answer:

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

Step-by-step explanation:

Data given and notation

$\bar X_{1}=1.17$ represent the mean for the sample 1 (25 mil film)

$\bar X_{2}=1.04$ represent the mean for the sample 2 (20 mil film)

$s_{1}=0.11$ represent the sample standard deviation for the sample 1

$s_{2}=0.09$ represent the sample standard deviation for the sample 2

$n_{1}=8$ sample size selected for 1

$n_{2}=8$ sample size selected for 2

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if reducing the film thickness increases the mean speed of the film, the system of hypothesis would be:

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

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

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

4 0

3 years ago

which ones belong to SAS, SSS, HL, AAS, ASA or not congruent?

slava [35]

Answer:

1.ASA

2.AAS

3.?

4.ASA

5.SSS

6.SSS

7.ASA

8 0

3 years ago

corinne is taking a train into the city. The train travels at a rate of 45 miles per hour. which quantity depends on the other?

lana66690 [7]

Complete Question:

Corinne is taking a train into the city. The train travels at a rate of 45 miles per hour. Which quantity depends on the other?

A. The distance traveled depends on the number of hours traveled.

B. The number of hours traveled depends on the speed of the train.

C. The speed of the train depends on the distance traveled.

D. The speed of the train depends on the number of hours traveled.

Answer:

Corinne is taking a train into the city. The train travels at a rate of 45 miles per hour. The distance traveled depends on the number of hours traveled

Step-by-step explanation:

As we know the velocity can de defined as the rate of change of position of objects in a particular time. The formula is given by,

$\text {velocity}=\frac{\text {distance}}{\text {time}}$

Here, in the question, rate of speed of train is given as 45 miles per hour. if we keep the velocity as constant one, then distance is directly proportionate to the travelled time

$\text {distance}=\text {velocity} \times \text {time}$