Which is a factor of 24x2 − 44x + 12

The dependent variable, y is 6 less than the independent variable, x.

Maslowich

D) because “y is” means “y equals” leaving only 2 choices, and it said “6 less than the independent variable, 6. (Y=6-x)

3 0

3 years ago

Which table represents a linear function?

Delvig [45]

Answer:

I would go with the first one.

Step-by-step explanation:

The first one is the only one that increases by the same number. In other words, its slope is constant. A linear function has a constant slope. You can tell that it increases by 1/2 every time.

6 0

3 years ago

Mon Tue Wed Thur Fri Sat Sun –2 –6 –3 2 6 –7 0 a. Arrange the temperatures in order from coldest to warmest. b. On a day in Janu

zvonat [6]

Answer:

A.) -7, -6, -3, -2, 0, 2, 6

B.) Jacob is correct, the larger the negative number, the lower the number is. So, since -11 is smaller than -7, it will be colder.

C.) Minneapolis is colder since -41 is way lower than -15.

-41<-15

7 0

3 years ago

Convert 290cm into m.

WARRIOR [948]

Answer:

2.90m

Step-by-step explanation:

I hope this is the answer.

3 0

2 years ago

Read 2 more answers

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