The points (3,9) and (-3,-9) are plotted on the coordinate plane using the equation y= a•x. What is the value of a?

Aaron gives an estimate of 2 for the quotient in a division problem his teacher says his estimate is reasonable if the divisor i

viva [34]

B because it's the closest to 8

6 0

3 years ago

How do I do this? Can somebody answer and explain?

yan [13]

A) multiple the 3 x 4 and add 1.
3 x 4 = 12 + 1 = 13
13/3

B) (7 x 3) + 2 = 23/3

C) (4 x 2) + 1 = 9/2

D) (2 x 8) + 7 = 23/8

5 0

2 years ago

Read 2 more answers

What is the product of 1/3 and 20.5?

tia_tia [17]

Answer:

6.83333333333 (repeated)

Step-by-step explanation:

1/3 x 20.5= 6.83333333333...

5 0

2 years ago

Read 2 more answers

PLZ PLZ PLZ PLZ PLZ PLZ PLZ PLZ PLZ HELP FOR BOTH OF THEM

tankabanditka [31]

Answer:

7) B, C

8) D

Step-by-Step-Explanation:

The first thing you need to solve for is the the parenthesis

the second one simplifies to x ≤ -7

4 0

2 years ago

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from

Vinvika [58]

Answer:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

$df=n_1 +n_2 -2=13+19-2=30$

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

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

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

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

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

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.

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

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

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

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

4 0

3 years ago