Solve the following inequality.<br> 6.37 <-45.36<br> Which graph shows the correct solution?

Which equation matches the graph?

tatiyna

Answer:

C

Step-by-step explanation:

C

7 0

3 years ago

Read 2 more answers

What is the average rate of change for the function ( IN THE PIC) over the interval (also in pic) HELP PLSSS ASAP!!!!

solong [7]

It is B because I had the question before in my school

3 0

3 years ago

Anna surveyed 300 of the students in her school about their favorite color. 186 students said their favorite color was red. What

MArishka [77]

Answer:

62%

Step-by-step explanation:

I took a fraction and made a decimal which was made into a percentage.

186 ÷ 300 = .62

Check: 62% • 300 = 186

6 0

3 years ago

A college job placement office collected data about students’ GPAs and the salaries they earned in their first jobs after gradua

frez [133]

Answer:

X is the GPA

Y is the Salary

Standard deviation of X is 0.4

Standard deviation of Y is 8500

E(X)=2.9

E(Y)=47200

We are given that The correlation between the two variables was r = 0.72

a) $y = a+bx$

$b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{r \times \sqrt{var(X) \times Var(Y)}}{Var(X)} = \frac{0.72 \times \sqrt{0.4^2 \times 8500^2}}{0.4^2} = 15300$

$a=y-bx = 47200-(15300 \times 29) = 2830$

So, slope = 15300

Intercept = 2830

So, equation : $y = 2830+15300x$

b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning?

$y = 2830+15300 \times 3.3 = 53320$

Observed salary = Residual + predicted = -1860+53320 = 51440

c)) What proportion of the variation in salaries is explained by variation in GPA?

The proportion of the variation in salaries is explained by variation in GPA = $r^2 = (0.72)^2 =0.5184$

8 0

3 years ago

The local swim team is considering offering a new semi-private class aimed at entry-level swimmers, but needs a minimum number o

SCORPION-xisa [38]

Answer:

The significance level is $\alpha=0.01$ and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

$z_{\alpha/2}= 2.326$

So we reject the null hypothesis is $z>2.326$

Step-by-step explanation:

For this case we define the random variable X as the number of entry-level swimmers and we are interested about the true population mean for this variable . On specific we want to test this:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

And the statistic is given by:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

The significance level is $\alpha=0.01$ and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

$z_{\alpha/2}= 2.326$

So we reject the null hypothesis is $z>2.326$

7 0

3 years ago

Solve the following inequality. 6.37 <-45.36 Which graph shows the correct solution?