Answer:
The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for the difference between population means is:
The information provided is as follows:
The critical value of <em>z</em> for 98% confidence level is,
Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:
Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).
Answer:
1
Step-by-step explanation:
divide the volume value by 16
<span>Table:
Class Boundaries Frequency
5-10 8
10-15 9
15-20 15
20-25 10
25-30 8
30-35 6
----------
total 56
Average =
[5+10]/2*8+[10+15]/2*9+[15+20]/2*15+[20+25]/2*10+[25+30]/2*8+[30+35]/2*6
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</span> 56
That is 1075 / 56 = 19.2
Answer: 19
X=price of one jumbo popcorn
y=price of one chocolate chip cookies
$5.00=$5.00(100 cts / $)=500 cts
$6.00=$6.00(100 cts / $)=600 cts
We suggest this system of equations:
x+2y=500
x+4y=600
we solve this system of equations by reduction method.
-(x+2y=500)
x+4y=600
----------------------
2y=100 ⇒ y=100/2=50
x+2y=500
x+2(50)=500
x+100=500
x=500-100
x=400
solution: one chocolate chip cookie cost 50 cts.
<span>Based on the data in the two-way table, the probability of being older than 25 years and having a hemoglobin level above 11 is
(154-69)/(429-139)
=85/290
=0.2931~0.29
Answer: A. 0.29
</span><span>The probability of having a hemoglobin level above 11 is
P(H>11)
=154/429
=0.35897~0.36
Answer: </span><span>C:0.36
</span><span>Being older than 25 years and having a hemoglobin level above 11
</span>Are not dependent on each other because w have not been told about any factors that were included in selection of sample.
Answer: <span>B.are not</span>
