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a_sh-v [17]
3 years ago
14

An athlete trains for 75 mins each day for as many days as possible write an equation that relates the number of days d that the

athlete spends training when the athlete trains for 375 mins
Mathematics
1 answer:
BartSMP [9]3 years ago
3 0
The answer would be"75d=375" because we know how many minutes in all and how many a minutes a day. so we need to figure how many days would it take
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In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were re
11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μemployed(μ1) be the sample mean of students who are employed and μnot employed(μ2) be the sample mean of students who are not employed

The random variable is μ1 - μ2 = difference in the mean of the employed and unemployed students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 < μ2 H1 : μ1 - μ2 < 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 3.22

μ2 = 3.33

s1 = 0.475

s2 = 0.524

n1 = 172

n2 = 116

t = (3.22 - 3.33)/√(0.475²/172 + 0.524²/116)

t = - 1.81

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.475²/172 + 0.524²/116]²/[(1/172 - 1)(0.475²/172)² + (1/116 - 1)(0.524²/116)²] = 0.00001353363/0.00000005878

df = 230

We would determine the probability value from the t test calculator. It becomes

p value = 0.036

Since alpha, 0.05 > than the p value, 0.036, then we would reject the null hypothesis.

Therefore, at a 5% significant level, this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed

3 0
3 years ago
Function h models an object's height in feet after x seconds have passed. Which equation shows that the object's height increase
ASHA 777 [7]

Answer:

C

Step-by-step explanation:

We know that function <em>h</em> represents an object's height in feet after <em>x</em> seconds.

In that case, option A) h(15) = 100 means that after 15 seconds, the object's height is 100 feet.

Option B) h(100) = 15 means that after 100 seconds, the object's height is 15 meters.

Therefore, neither A nor B are correct.

Option C) h(15) - h(0) = 100 means that between the zeroth and 15th second, their difference is 100 feet.

In other words, the object's height increased by 100 feet over the first 15-second period.

Option C is correct.

For Option D), it gives us the average rate of change. (h(15) - h(0)) / (15) = 100 means that for the first fifteen seconds, the height of the object increased at an average rate of 100 feet per second.

8 0
3 years ago
Jkhhhhljhjklhkljhkljhkljhjkhjklhjk
Vsevolod [243]

Answer:

hi

Step-by-step explanation:

6 0
3 years ago
Determine if x + 2 is a factor of p(x) = x ^4 + 3x^ 3 + 4x ^2 - 8 and explain why.
zlopas [31]
X^4+ 3x^3+4x^2-8 |   x+2   =  x^3+x^2+2x-4
 -x^4-2x^3
-----------------------     
         x^3+4x^2-8
         -x^3-2x^2
-------------------------
                 2x^2-8
                 -2x^2-4x
                --------------
                  -4x-8
                    4x+8
----------------------------
                      / /

is a factor because (x+2)(x^3+x^2+2x-4)=x^4+3x^3+4x^2-8
4 0
3 years ago
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For a experimental to produce useful data, what must happen
kozerog [31]

Answer:

cr qe deobpaase r

Step-by-step explanation:

5 0
3 years ago
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