True or False

seropon [69]

Answer:
8x+y

Explanation:
3x+4y+5x-3y
Rearrange the terms so like terms are next to each other
3x+5x+4y-3y
Combine like terms
8x+y

I hope this helps!

7 0

3 years ago

The mean finish time for a yearly amateur auto race was 186.94 minutes with a standard deviation of 0.372 minute. The winning ca

Fed [463]

Answer:

Sam had a finish time with a z-score of -2.93.

Karen had a finish time with a z-score -1.91.

Sam had the more convincing victory.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Sam:

The mean finish time for a yearly amateur auto race was 186.94 minutes with a standard deviation of 0.372 minute. The winning car, driven by Sam, finished in 185.85 minutes.

So $\mu = 186.94, \mu = 0.372, X = 185.85$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{185.85 - 186.94}{0.372}$

$Z = -2.93$

Sam finishing time was 2.93 standard deviations below the mean.

Sam had a finish time with a z-score of -2.93.

Karen:

The previous year's race had a mean finishing time of 110.7 with a standard deviation of 0.115 minute. The winning car that year, driven by Karen, finished in 110.48 minutes.

So $\mu = 110.7, \sigma = 0.115, X = 110.48$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{110.48 - 110.7}{0.115}$

$Z = -1.91$

Karen finishing time was 1.91 standard deviations below the mean.

Karen had a finish time with a z-score -1.91.

Who had the more convincing victory?

Sam finishing time was 2.93 standard deviations below the mean.

Karen finishing time was 1.91 standard deviations below the mean.

Sam finished more standard deviations below the mean than Karen, that is, he was faster relative to his competition than Karen.

So Sam had the more convincing victory.

8 0

4 years ago

Which one is it? 1,2,3 or 4?

Kobotan [32]

Answer:

3

Step-by-step explanation:

If you look on the graph, you can see that the y intercept is at -4. This easily takes out answers 1 and 2 because they have a y intercept a +3. Next you notice that any point with a y value must be equal to or less than 4/3x-4. Since every point below the line has a y value less than any point on the line, it must mean that y< or =4/3x-4

7 0

3 years ago

1000 grams<br> 1 kilogram<br> 6,000 grams<br> x kilograms

Dominik [7]

Answer:

6 kilograms

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A sample of 41 observations is selected from one population with a population standard deviation of 3.4. The sample mean is 100.

Marizza181 [45]

Answer:

a) If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

b) (-1.9886, 1.9886)

c) $t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617$

d) $p_v =2*P(t_{84}>1.617) =0.1096$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.

e) $p_v =2*P(t_{84}>1.617) =0.1096$

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

Our notation on this case :

$n_1 =41$ represent the sample size for group 1

$n_2 =45$ represent the sample size for group 2

$\bar X_1 =100$ represent the sample mean for the group 1

$\bar X_2 =98.4$ represent the sample mean for the group 2

$s_1=3.4$ represent the sample standard deviation for group 1

$s_2=5.6$ represent the sample standard deviation for group 2

Part a

If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

Part b

On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution $\alpha/2 = 0.025$ of the area.

The distribution on this cas since we don't know the population deviation for both samples is the t distribution with $df=n_1+n_2 -2= 41+45-2=84$ degrees of freedom.

We can use the following excel codes in order to find the critical values:

"=T.INV(0.025,84)", "=T.INV(1-0.025,84)"

And we got: (-1.9886, 1.9886)

Part c

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

And now we can calculate the statistic:

$t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617$

The degrees of freedom are given by:

$df=41+45-2=84$

Part d

And now we can calculate the p value using the altenative hypothesis:

$p_v =2*P(t_{84}>1.617) =0.1096$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.

Part e

$p_v =2*P(t_{84}>1.617) =0.1096$

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

8 0

3 years ago