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Roman55 [17]
3 years ago
8

Having done poorly on their math final exams in​ June, six students repeat the course in summer​ school, then take another exam

in August. If we consider these students representative of all students who might attend this summer school in other​ years. Do these results provide evidence that the program is worthwhile?
June 54, 49, 68, 66, 62, 62

August 50, 65, 74, 64, 68, 72

Required:
Evaluate on a TI 83 calculator
Mathematics
1 answer:
pochemuha3 years ago
6 0

Answer:

Yo yo

Step-by-step explanation:

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ale4655 [162]

Answer:

x = 44º

Step-by-step explanation:

∡P  = 360 - 112*2 = 136º

∡A = x = 180 - 136º = 44º

6 0
3 years ago
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The sum of two numbers is 64 and their differences in 10. Find the numbers
uranmaximum [27]

Answer:

37 and 27

Step-by-step explanation:

Lets say that the two numbers are x and y. We get the equations x + y = 64 and x - y = 10. We can then add these two equations together to get, 2x = 74. Then we divide by two on both sides and get x = 37. We then plug this value back into the second equation and get that 37 - y = 10. We can simplify this and we get that y = 27. So our two numbers are 37 and 27.

4 0
2 years ago
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What is the mean in math?
sertanlavr [38]
Adding all vakus in data set and divide by the number of values in the data set
3 0
3 years ago
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Solve exponential equation<br> 1/16=64^4x-3
RoseWind [281]
Fraction form: x=7/12

Steps:

64^4x-3

2^6(4x-3)=2^-4

6(4x-3)=-4

4x-3=-2/3

4x=7/3

X=7/12
4 0
3 years ago
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A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student
umka2103 [35]

Answer:

(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908  

(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308  

So then the confidence interval for the difference of means is given by:

-8.908 \leq \mu_M -\mu_F \leq 5.308

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X_{M}= 69.8 represent the mean for the age of grandmother

\bar X_{F}= 71.6 represent the mean for the age of grandfather

s_{M}= 8.38 represent the sample deviation for the age of grandmother

s_{F}= 6.65 represent the sample deviation for the age of grandfather

n_M = n_F= 10

Solution to the problem

For this case the confidence interval is given by:

(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n_M +n_F-2=10+10-2=18

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that t_{\alpha/2}=2.101

And replacing we got:

(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908  

(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308  

So then the confidence interval for the difference of means is given by:

-8.908 \leq \mu_M -\mu_F \leq 5.308

5 0
4 years ago
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