Please can someone help me

Read the picture above

Romashka-Z-Leto [24]

Vitvivuc vrivdfdxfdddxgtdcgyimbfdd

5 0

3 years ago

Tres fracciones que representen el 3

FrozenT [24]

Creo que las respuestas son 6/2, 18/6 y 12/4

7 0

3 years ago

Read 2 more answers

company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of attaining a fleet average of at least 26

podryga [215]

Answer:

$t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435$

$p_v =P(t_{(49)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Step-by-step explanation:

Data given and notation

$\bar X=25.02$ represent the sample mean

$s=4.83$ represent the sample standard deviation

$n=50$ sample size

$\mu_o =26$ represent the value that we want to test

$\alpha$ 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 the mean is at least 26 mpg, the system of hypothesis would be:

Null hypothesis: $\mu \geq 26$

Alternative hypothesis: $\mu < 26$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435$

P-value

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

$df=n-1=50-1=49$

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

$p_v =P(t_{(49)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

3 0

3 years ago

Find a value of k for which the following system has only one solution:

densk [106]

K(cannot equal (the = sign with the line through it) -2
k(=/)-2

4 0

3 years ago

Surya is competing in a triathlon that includes swimming, bicycling and running. For the first two segments of the race, Surya s

Alex73 [517]

Step-by-step explanation:

Let $d_{1}$ be the distance Surya swam and $d_{2}$ be the distance he bicycled. We know that $d_{1} + d_{2}$ = 43.29 km.

Now let $t_{1}$ and $t_{2}$ be the time, in hours, it took Surya to complete the swimming and bicycling portions of the race, respectively.

That means $t_{1} + t_{2}$ = 81.75/60 = 1.3625 hours, and $t_{2}$ = 1.3625 – $t_{1}$

Using the distance formula, we have $d_{1}$ = 4 $t_{1}$ and $d_{2}$ = 40 $t_{2}$

= 40(1.3625 – $t_{1}$ )

= 54.5 – 40 $t_{1}$

Substituting these expressions for $d_{1}$ and $d_{2}$ in our first equation yields 4 $t_{1}$ + 54.5 – 40 $t_{1}$ = 43.29.

Solving for $t_{1}$ we see that - 36 $t_{1}$ = - 11.21, meaning it took Surya $t_{1}$ = 0.31138(Bar) hours to complete the swimming portion of the race.

This is equivalent to 0.31138(Bar)(60) = 18.683(Bar)≈ 18.68 minutes.

8 0

3 years ago