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Anni [7]
3 years ago
14

Jason earned scores of 90 points, 82 points, 95 points, and 86 points on four tests. What is the lowest score he can get on the

fifth test and still finish with an average score of 90 points?
95 points
96 points
97 points
98 points
Mathematics
2 answers:
dem82 [27]3 years ago
7 0
97 points is your answer 3 hope this helps!
Aleonysh [2.5K]3 years ago
5 0

Answer:

C 97

Step-by-step explanation:

So you know that average is equal to all the scores added together divided by the amount of tests right? Now you have 5 tests and you need an average of 90 points. Average=scores added together/no. of test

90=x/5

90*5=x

450=x

Now take 450 subtract all your other 4 scores to get your lowest 5th score

450-90-82-95-86=97

You need at least 97 to have an average of 90.

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An employer wishes to compare typing speeds of graduates from 2 different study programs: A and B. Eight random graduates from c
Karolina [17]

Answer:

Calculated value t =  0.614< 1.782

we accepted null hypothesis for 12 degrees of freedom at 0.05 level of significance .

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

<u>we will t-test </u> t = x⁻- y⁻ /S√ n₁+n₂-2

Given An employer wishes to compare typing speeds of graduates from 2 different study programs: A and B

course A type at 62, 85, 59, 64, 73, 70, 75, and 72

mean of x is x⁻ = ∑x / n

mean of x = 62+85+59+64+73+70+75+ 72 /8 = 70

x⁻ = 70

course B type at 75, 64, 81, 55, 69, and 58

mean of y is  y⁻ = ∑y / n = 75+64+81+55+69+ 58/6 = 67

y⁻ =67

course A    course B

    x                    y             x- x⁻     (x-x⁻ )^2     y- y⁻     (y-y⁻ )^2    

    62                75            8             64          8            64

    85                64           15            225       -3             9

    59                81           -11            121          14            196      

    64                55           -6            36          -12          144          

    73                69          3                9            2            4

    70               58            0              0           -9            81

    75                              5              25

    72                              2               4

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Adding ∑( (x-x⁻ )^2+ ∑( (y-y⁻ )^2= 982

n₁+n₂-2 = 8+6-2=12

now S^2 =  982/ 12 =81.833

<u>Null hypothesis :H₀:μ₁=μ₂ ( there is no difference between A and B)</u>

<u>Alternative hypothesis :H₁:μ₁≠μ₂</u>

<u>level of significance ∝=0.05</u>

<u>we will use t-test </u> t =   0.614 ( see attachment )

The degrees of freedom = n₁+n₂-2 = 8+6-2=12

From tabulated value  t = 1.782

Calculated value t = 0.614< 1.782

we accepted null hypothesis for 12 degrees of freedom at 0.05 level of significance .

There is no difference between Course A and Course B

Download docx
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