In which quadrant(s) are tan(x) and sin(x) both negative? quadrant II quadrants II and III quadrant IV quadrants III and IV

20.155 to the nearest ten

Radda [10]

Answer:

20.2

Step-by-step explanation:

because its 20.2 trust

plz mark brainliest

4 0

3 years ago

The rectangle has an area of x2+13x+36 square meters and a length of x+9 meters.

Brut [27]

Answer:The answer is 58

Step-by-step explanation:

4 0

4 years ago

A soda can has a diameter of 6 cm. What is the circumference of the can?

nexus9112 [7]

C = 2PIr

C = 2 * 3.14 * 3 = 18.84 cm

7 0

4 years ago

5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim

Ne4ueva [31]

Answer:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

$S_p=2.940$

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

$df=60+72-2=130$

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

$n_1 =60$ represent the sample size for people who used a self service

$n_2 =72$ represent the sample size for people who used a cashier

$\bar X_1 =5.2$ represent the sample mean for people who used a self service

$\bar X_2 =6.1$ represent the sample mean people who used a cashier

$s_1=3.1$ represent the sample standard deviation for people who used a self service

$s_2=2.8$ represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

And t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

System of hypothesis

Null hypothesis: $\mu_1 \geq \mu_2$

Alternative hypothesis: $\mu_1 < \mu_2$

This system is equivalent to:

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

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

We can find the pooled variance:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

And the deviation would be just the square root of the variance:

$S_p=2.940$

The statistic is given by:

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

The degrees of freedom are given by:

$df=60+72-2=130$

And now we can calculate the p value with:

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

5 0

3 years ago

Pls help with my geometry

Sati [7]

Least to greatest is mB,mD,mC

8 0

3 years ago