How would we solve 6/15=2/c

Answers? & provide how u did it if u can thank u.

sp2606 [1]

Answer:

8

Step-by-step explanation:

a² = 10² - 6²

a² = 100 - 36

a² = 64

a = √64

a = 8

So, the value of a is 8

Hope this is helpful and useful :)

7 0

3 years ago

Read 2 more answers

Help please:))) i just wanna pass:/

OleMash [197]

Answer:

1,3,4 I believe is correct

4 0

3 years ago

X + 2 y is greater than or equal to negative 3

EastWind [94]

X+2y $\geq$ -3 that may be wrong

7 0

4 years ago

The TV screen is similar to the computer screen. What is the area of the TV screen?

tigry1 [53]

12 not sure if it's true or not.

8 0

3 years ago

A student at a four-year college claims that average enrollment at four-year colleges is higher than at two-year colleges in the

Kitty [74]

Answer:

a) Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

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

The degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$ Replacing the info given we got:

$t=\frac{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921$

c) $p_v =P(t_{68}>0.0921)=0.463$

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean of enrollment is significantly higher for four year college than for two year college

Step-by-step explanation:

Information given

$\bar X_{1}=5069$ represent the mean for two year colleges

$\bar X_{2}=5216$ represent the mean for four year college

$s_{1}=4773$ represent the sample standard deviation for 1

$s_{2}=8141$ represent the sample standard deviation for 2

$n_{1}=35$ sample size for the group 2

$n_{2}=35$ sample size for the group 2

$\alpha=0.05$ Significance level provided

Part a

Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

Part b

We want to test if the mean of enrollment for four year college is higher than for two year college, the system of hypothesis would be:

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

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

The statistic is given by:

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

The degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$ Replacing the info given we got:

$t=\frac{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921$

Part c

The p value would be given by:

$p_v =P(t_{68}>0.0921)=0.463$

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean of enrollment is significantly higher for four year college than for two year college

6 0

4 years ago