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Vesnalui [34]
3 years ago
15

Please please help with this question , you will get 10 points if you answer this question, don’t get bothered with 1 point writ

ten in question.

Mathematics
2 answers:
bazaltina [42]3 years ago
6 0

Answer:

11.7

Step-by-step explanation:

pls follow me and !ark as brainliest and follow me to get free points to all your answer's

faltersainse [42]3 years ago
5 0

Answer:

B (11.7)

Step-by-step explanation:

if you add the whole numbers (4, 4, 3) it would be 11 and then you add the decimal part which is 7 so it would be 11.7

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An instructor made a frequency table of the scores his students got on a test score frequency 30-under 40 1 40-under 50 4 50-und
RideAnS [48]
First, you add up the total number of students because we would base our percentage one this.

Total number of students = 1+4+5+10+20+10+5
Total number of students = 55

Now, let's count how many of them got a score of more than 70. For this, you would only count the students belonging to 70-under 80, 80-under 90 and 90-under 100.

Students who got more than 70 = 20+ 10 + 5 = 35

Therefore the percentage is equal to:
Percentage = 35/55 * 100 = 63.64%
4 0
3 years ago
What is the solution to the equation-3(h+5)+2 = 4(h+6)- 9?<br> h= 4<br> h= -2<br> h = 2<br> h = 4
ruslelena [56]
Expanding
-3h-15+2=4h+24-9
-3h-4h=24-9-2+15
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h=4
8 0
2 years ago
Find the domain of the graphed function.
attashe74 [19]

Answer:

b

Step-by-step explanation:

7 0
2 years ago
Please do Correctly will mark you Brainliest!
ivolga24 [154]

Answer:

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4 0
3 years ago
Read 2 more answers
Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar
klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack

Where (from tables):

Z_{0.99}=2.33

Finally, the interval at 98% confidence level is:

CI(\mu)=\lbrack28.94,31.06\rbrack

4 0
1 year ago
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