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ycow [4]
3 years ago
15

A random sample of 120 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviatio

n of 240. If we want to provide a 95% confidence interval for the SAT scores, the degrees of freedom for reading the critical values of the "t" value is:_________
Mathematics
1 answer:
sammy [17]3 years ago
8 0

Answer:

1.980

Step-by-step explanation:

Given that a random sample of 120 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240

Here nothing mentioned about the distribution but sample size is very large.

Since population std deviation is not known, only sample std deviation is used.

t test will be more appropriate.

df = degrees of freedom = n-1=119

For two tailed 95% level, t critical value for 119 df would be

1.980

Using this critical value confidence interval would be formed.

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Solve the equation for y:<br> 2x + 5y = 20
lbvjy [14]

Answer:

y = \frac{20-2x}{5}

Step-by-step explanation:

Given

2x + 5y = 20 ( isolate the term in y by subtracting 2x from both sides )

5y = 20 - 2x ( divide both sides by 5 )

y = \frac{20-2x}{5}

6 0
3 years ago
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McKenzie runs 1.5 mi every 10 min.
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If McKenzie runs 1.5 mp10m, multiply 1.5 by 6. McKenzie runs 9 miles per hour.
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3 years ago
the diameter of a piston for an automobile is 3 11/16in with tolerence of 1/64in. find the upper and lower limits of the pistons
padilas [110]

Upper Tolerance

Remark

The 11/16 is the only thing that will be affected. The three won't go up or down when we add 1/64 so we should just work with the 11/16. We need only add 11/16 and 1/64 together to see what the upper range is. Later on we can add 3 into the mix.

Solution

<u>Upper Limit</u>

\dfrac{11}{16} +  \dfrac{1}{64}

Now change the 11/16 into 64. Multiply numerator and denominator or 11/16 by 4

\dfrac{11*4}{16*4} +  \dfrac{1}{64}

\dfrac{11*4}{16*4} +  \dfrac{1}{64} Which results in

\dfrac{44}{64} +  \dfrac{1}{64}

With a final result for the fractions of 45/64

So the upper tolerance = 3 45/64

<u>Lower Tolerance</u>

Just follow the same steps as you did for the upper tolerance except you subtract 1/64 like this.

\dfrac{11}{16} - \dfrac{1}{64}

Your answer should be 3 and 43/64


4 0
3 years ago
State the open intervals over which the function is (a) increasing, (b) decreasing or (c) constant
ipn [44]

Answer:

a. [-3, 4]

b. (-inf, -3]

c. [4, inf)

Step-by-step explanation:

Our intervals will represent the x-values

We know that since there's an arrow pointing to the left of the line that it goes on infinitely

Same thing when the arrow is going to the right

Then we can just looking at the x-values on the graph for the intervals where it starts and stops

Hope this helps

Best of luck

3 0
2 years ago
743 times 295 answer please
Mamont248 [21]
The answer is 219,775
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3 years ago
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