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Mama L [17]
4 years ago
8

When a class took a math test, 15% of the class failed, 25% made some mistakes (but didn’t fail), and 24 students got perfect sc

ores. How many students were in the class?
Mathematics
2 answers:
gavmur [86]4 years ago
7 0
In the class there would be a total of 40 students
Degger [83]4 years ago
4 0
15%+25%=40%, meaning that 40% of the class didn’t get a perfect score and 60% did. If 24 students got a perfect score and 24 students is equal to 60% of the class, you can solve by setting up a proportion.

24 students X students
——————- = —————-
60% of class 100% of class

Next, cross multiply to get 2400 = 60X. Divide both sides by 60 and you’ll get 40. There are 40 students in the class.


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To get to work, you drive for 20 minutes at 30 mph and walk 10 more minutes at 3 mph What was your average speed?
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A bank is trying to determine which model of safe to install. The bank manager believes that each model is equally resistant to
alexandr402 [8]

Answer:

So on this case the 95% confidence interval would be given by (-1.152;5.152).

-1.152 < \mu_{safe1}-\mu_{safe2}  

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for ?D contains 0.

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".  

Solution

Let put some notation  

x=value for Safe 1 , y = value for Safe 2

x: 103, 90, 64, 120, 104, 92, 145, 106, 76

y: 101, 94, 58, 112, 103, 90, 140, 110, 74

The first step is calculate the difference d_i=x_i-y_i and we obtain this:

d: 2, -4, 6, 8, 1, 2, 5, -4, 2

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{18}{9}=2

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =4.093

The next step is calculate the degrees of freedom given by:

df=n-1=9-1=8

Now we need to calculate the critical value on the t distribution with 8 degrees of freedom. The value of \alpha=1-0.95=0.05 and \alpha/2=0.025, so we need a quantile that accumulates on each tail of the t distribution 0.025 of the area.

We can use the following excel code to find it:"=T.INV(0.025;8)" or "=T.INV(1-0.025;8)". And we got t_{\alpha/2}=\pm 2.31

The confidence interval for the mean is given by the following formula:  

\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

Now we have everything in order to replace into formula (1):  

2-2.31\frac{4.093}{\sqrt{9}}=-1.152  

2+2.31\frac{4.093}{\sqrt{9}}=5.152  

So on this case the 95% confidence interval would be given by (-1.152;5.152).

-1.152 < \mu_{safe1}-\mu_{safe2}  

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for D contains 0.

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3 years ago
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Vaselesa [24]

Answer:

p = 56\% = 0.56 \\ n = 5 \\ q = 1 - p = 0.44 \\ P(X = x) =  {n \choose x}  {p}^{x}  {q}^{(n - x) }  \\ P(X < 2) = P(X = 0) + P(X = 1) \\  {5 \choose 0}  {0.56}^{0}  {0.44}^{(5 - 0) } +  {5 \choose 1}  {0.56}^{1}  {0.44}^{(5 - 1) }

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3 years ago
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Mkey [24]

Answer: B) A = 750(1.04)ⁿ

<u>Step-by-step explanation:</u>

The formula for compounded annually is: A = P(1 + r)ⁿ   where

  • A (amount accrued) = <em>unknown</em>
  • P (amount invested) = $750
  • r (interest rate) = 4% -->(0.04)
  • t (time in years) = <em>unknown</em>

A = 750(1 + 0.04)ⁿ

  = 750(1.04)ⁿ

8 0
3 years ago
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