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jeka57 [31]
3 years ago
5

Recently, Frank took a one-hundred question aptitude test where each correct answer scored 5 points, each incorrect answer score

d -2 points, and each question not answered scored zero points. Frank answered 80 questions and scored 232 points. How many questions did he answer correctly?
Mathematics
2 answers:
Neporo4naja [7]3 years ago
8 0

Answer:

56 questions were correctly answered by Frank.

Step-by-step explanation:

Let Frank answered the number of correct questions = x

Let wrong answers given by Frank = y

By statement 1 "Frank answered 80 questions".

Equation will be x + y = 80 ---------(1)

Statement 2 "Frank scored 232 points."

equation will be 5x - 2y = 232 ------(2)

We have to calculate number of correct answers.

Multiply equation (1) by (2) then add it to equation (1).

(2x + 2y) + (5x - 2y) = 160 + 232

7x = 392

x = 56

Therefore, 56 questions were correctly answered by Frank.

Lyrx [107]3 years ago
3 0
Im guessing he answered 30 correct, let me know if im correct
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5 0
4 years ago
Read 2 more answers
The national average for the number of students per teacher for all U.S. public schools in 15.9. A random sample of 12 school di
Nady [450]

Answer:

The 95% confidence interval for the true mean number of students per teacher is (17.9, 20.5). The lower bound of the interval is above the national average, which means that this estimate is higher than the national average.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 12 - 1 = 11

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 11 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.201

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.201\frac{\sqrt{4.41}}{\sqrt{12}} = 1.3

In which s is the standard deviation of the sample(square root of the variance) and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 19.2 - 1.3 = 17.9 students.

The upper end of the interval is the sample mean added to M. So it is 19.2 + 1.3 = 20.5 students

The 95% confidence interval for the true mean number of students per teacher is (17.9, 20.5). The lower bound of the interval is above the national average, which means that this estimate is higher than the national average.

5 0
3 years ago
A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observationa
diamong [38]

Answer:

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}= \mu_D

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C,D

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have p=4 groups and on each group from j=1,\dots,p we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2  

And we have this property  

SST=SS_{between}+SS_{within}  

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.

8 0
4 years ago
The vertices of quadrilateral MNPQ are M(−3,−2),N(−1,4),P(2,4), and Q(4,−2). Translate quadrilateral MNPQ using the vector ⟨3,−4
muminat

Answer:

see explanation

Step-by-step explanation:

A translation using the vector < 3, - 4 > , means

Add 3 to the x- coordinate and subtract 4 from the y- coordinate, so

M (- 3, - 2 ) → M' (- 3 + 3, - 2 - 4 ) → M' (0, - 6 )

N (- 1, 4 ) → N' (- 1 + 3, 4 - 4 ) → N' (2, 0 )

P (2, 4 ) → P' (2 + 3, 4 - 4 ) → P' (5, 0 )

Q (4, - 2 ) → Q' (4 + 3, - 2 - 4 ) → (7, - 6 )

4 0
3 years ago
If you plot and connect these points P1(3,2) and P2(6,8) , what figure is Formed? What is the Direction?
Sergio [31]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the points

  • P1(3, 2)
  • P2(6, 8)

When we plot P1(3, 2) and P2(6, 8), we determine the line segment P1P2.

The direction of the segment is from P1 to P2.

Determining the length of the segment P1P2.

\:l=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

  =\sqrt{\left(6-3\right)^2+\left(8-2\right)^2}

  =\sqrt{3^2+6^2}

  =\sqrt{45}

  =\sqrt{5}\sqrt{3^2}

  =3\sqrt{5}

Thus, the length of the segment is:

\:\:\:l=3\sqrt{5}

Determining the equation of a line containing the segment P1P2

Given the points

  • P1(3, 2)
  • P2(6, 8)

Finding the slope between P1(3, 2) and P2(6, 8)

\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}

\left(x_1,\:y_1\right)=\left(3,\:2\right),\:\left(x_2,\:y_2\right)=\left(6,\:8\right)

m=\frac{8-2}{6-3}

m=2

Using the line point-slope form of the line equation

y-y_1=m\left(x-x_1\right)

where m is the slope of the line and (x₁, y₁) is the point

substituting the values m = 2 and the point (3, 2)

y-2=2\left(x-3\right)

Add 2 to both sides

y-2+2=2\left(x-3\right)+2

y=2x-4

Thus, the equation of a line containing the line segment P1P2 is:

y=2x-4

8 0
3 years ago
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