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IgorLugansk [536]

2 years ago

7

The manager of a grocery store selected a random sample of 100 customers to estimate the average checkout time. The 90% confiden

ce interval estimate for the average checkout time at the score is[1, 5] minutes. The correct confidence interval interpretation is ___________.
a. 100.0 to 200.0
b. If random samples of 100 customers are repeatedly selected, then that the average checkout time will be between 1 and 5 minutes for 95% of the samples.
c. We estimate with 95% confidence that the average checkout time for the sample of 100 customers is between 1 and 5 minutes.
d. None of these alternatives is correct.
e. We estimate with 95% probability that the average checkout time at the store is between 1 and 5 minutes.

1 answer:

lina2011 [118]2 years ago

3 0

Answer:

c

Step-by-step explanation:

You might be interested in

Graph the system of equations on your graph paper to answer the question.

scZoUnD [109]

<span>y=x-4
y=-x+6
---------------------
Substitute x - 4 for y in </span>y=-x+6
x-4=-x+6<span>
--------------------------------
Add X on each side
</span><span>x-4+x</span>=-<span>x+6+x
</span>2x - 4 = 6
--------------------------------------
Add 4 on each side
<span>2x-4+4</span>=<span>6+4
</span>2x = 10
------------------------------------------------
Divide by 2 on each side
2x ÷ 2 = 10 ÷ 2
x = 5
Now we have X
================================================================
To find Y we substitute 5 for x in y=<span>x-<span>4
</span></span>y = 5 - 4
y = 1
-----------------------------------------------------
Your answers are Y = 1 and X = 5

8 0

2 years ago

Can someone help pls 18-3 X 2/5-1^2

d1i1m1o1n [39]

Answer:

3

Step-by-step explanation:

Remember the order of operations using PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). So, you get 18 - 6 / 5-1 = 12/4 = 3

7 0

2 years ago

Read 2 more answers

Help me it due 30 minutes <br> Estimate each product <br><br> 1/5x22

Alchen [17]

Answer:

I say its 4.4

Step-by-step explanation:

3 0

2 years ago

Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution. 2x-y=

Bond [772]

Answer:

The system of linear equations has infinitely many solutions

Step-by-step explanation:

Let's modified the equations and find the answer.

Using the first equation:

$2x-y=5$ we can multiply by 2 in both sides, obtaining:

$2*(2x-y)=2*5$ which can by simplified as:

$4x-2y=10$ which is equal to:

$4x=2y+10$

Considering the second equation:

$=4x+ky=2$

Taking into account that from the first equation we know that: $4x=2y+10$ , we can express the second equation as:

$2y+10+ky=2$ , which can be simplified as:

$(2+k)y=2-10$

$(2+k)y=-8$

$y=-8/(2+k)$

Because (-8) is being divided by (2+k), then (2+k) can't be equal to 0, so:

$2+k=0$ if $k=-2$

This means that k can be any number different than -2, and for each of these solutions, there is a different solution for y, allowing also, different solutions for x.

For example, if k=0 then

$y=-8/(2+0)$ which give us y=-4, and, because:

$4x=2y+10$ if y=-4 then $x=(-8+10)/4=0.5$

Now let's try with k=-1, then:

$y=-8/(2-1)$ which give us y=-8, and, because:

$4x=2y+10$ if y=-8 then $x=(-16+10)/4=-1.5$ .

Then, the system of linear equations has infinitely many solutions

8 0

2 years ago

The values of 60, 62, and 84 were common to both samples. The three values are identified as outliers with respect to the age-gr

postnew [5]

Answer:

Only 60 is an identified outlier

Step-by-step explanation:

Given

<u>Age group: 40 to 50</u>

$Min = 60$ $Q_1 = 70$ $Median= 73$

$Q_3 = 76$ $Max = 85$

<u>Conditions for outlier</u>

$Value > Q_3 + 1.5 * IQR$

$Value < Q_1 - 1.5 * IQR$

<em>See attachment for complete question</em>

<u>Required</u>

Which of 60, 62 and 84 is an outlier of age grout 40 to 50 years

First, calculate the IQR of group 40 to 50.

This is calculated as:

$IQR = Q_3 - Q_1$

Where:

$Q_3 = 76$ and $Q_1 = 70$

So:

$IQR = 76 - 70$

$IQR = 6$

<em>Next, is to test the outlier conditions on each value (i.e. 60, 62 and 84)</em>

<u></u>

<u>Testing 60</u>

<u>Condition 1</u>

$Value > Q_3 + 1.5 * IQR$

$60 > 76 + 1.5 * 6$

$60 > 76 + 9$

$60 > 85$ --- False

<u>Condition 2</u>

$Value < Q_1 - 1.5 * IQR$

$60 < 70 - 1.5 * 6$

$60 < 70 - 9$

$60 < 61$ --- True

<em>Because one of the conditions is true, then 60 is an outlier of group 40 - 50 years</em>

<em></em>

<u>Testing 62</u>

<u>Condition 1</u>

$Value > Q_3 + 1.5 * IQR$

$62> 76 + 1.5 * 6$

$62 > 76 + 9$

$62 > 85$ --- False

<u>Condition 2</u>

$Value < Q_1 - 1.5 * IQR$

$62 < 70 - 1.5 * 6$

$62 < 70 - 9$

$62 < 61$ --- False

<em>Because both conditions are false, then 62 is not an outlier of group 40 - 50 years</em>

<em></em>

<u>Testing 84</u>

<u>Condition 1</u>

$Value > Q_3 + 1.5 * IQR$

$84> 76 + 1.5 * 6$

$84 > 76 + 9$

$84 > 85$ --- False

<u>Condition 2</u>

$Value < Q_1 - 1.5 * IQR$

$84 < 70 - 1.5 * 6$

$84 < 70 - 9$

$84 < 61$ --- False

<em>Because both conditions are false, then 84 is not an outlier of group 40 - 50 years</em>

5 0

2 years ago