PLS HELP! (THE QUESTION IS A PICTURE CLICK ON THE QUESTION FIRST)

There are 80 people in Ms. Peterson's classes. One-fourth of the students are going on a field trip. Of those students, one-half

Naily [24]

20 students are going to the trip and of those 20, 10 are bringing their own lunches

3 0

3 years ago

9+15÷3(2)-6<br> Please help me

Firlakuza [10]

Answer:

13

Step-by-step explanation:

4 0

2 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

Put in your best Joke for free points

Fofino [41]

What do you call a bus full of white people? A Twinkie

4 0

3 years ago

Read 2 more answers

1. Write a rule that will take the triangle to a congruent triangle. Write your rule as (x,y)

olga nikolaevna [1]

Transforming a shape involves changing the size and/or the location of the shape.

The rule that will take the triangle to a congruent triangle is $(x,y) \to (x,-y)$
The rule that will take the triangle to a figure that is similar is $(x,y) \to (0.5x,0.5y)$ .
The rule that will take the triangle to a figure that is not similar or congruent is $(x,y) \to (0.5x,2y)$ .

<h3>Rigid transformation</h3>

When a shape is transformed by a rigid transformation, then the image of the shape will be congruent to the original shape.

An example of a rigid transformation is a reflection over the x-axis; and the rule is:

$(x,y) \to (x,-y)$

So, a rule that will take the triangle to a congruent triangle is $(x,y) \to (x,-y)$

<h3>Nonrigid transformation</h3>

When a shape is transformed by a nonrigid transformation, then the image of the shape will not be congruent to the original shape, however the original shape and the transformed shape would be similar

An example of a nonrigid transformation is a dilation by a scale factor od 0.5; and the rule is:

$(x,y) \to (0.5x,0.5y)$

So, a rule that will take the triangle to a figure that is similar is $(x,y) \to (0.5x,0.5y)$

However, if the x and y coordinates of the shape are dilated by different scale factors, then the resulting shape would neither be similar nor be congruent to the original shape.

A rule that will take the triangle to a figure that is not similar or congruent is $(x,y) \to (0.5x,2y)$ .