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RUDIKE [14]
3 years ago
9

Solve for t in terms of s, u, and v

Mathematics
1 answer:
e-lub [12.9K]3 years ago
7 0

Answer:

t = - \frac{uv}{s}

Step-by-step explanation:

multiply both sides by u to eliminate the fraction

uv = - st

isolate t by dividing both sides by - s

- \frac{uv}{s} = t


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What is 185/100+50%+4%
Natali5045456 [20]

Answer:

2.815

Step-by-step explanation:

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3 years ago
What is the y-intercept of the line given by the equation below ? Y=8x+7
kow [346]
The y-intercept is 7.

The equation is in form y = mx + b.

b is the y-intercept and m is the slope.

So in this case the slope is 8.

Hope this helps :)
7 0
3 years ago
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Lucy weighed an object on her balance and recorded amass of 36.5 grams . the weight on the label says it should . weigh 36.8 gra
Lunna [17]
Error in the weight is 0.3g (Difference between the actual 36.8g and the experimented 36.5g)
So %Error = Error/Correct value *100%
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4 0
3 years ago
Which values will complete the table.<br> A<br> B<br> C
Anvisha [2.4K]

Answer:

A=8

b=15

c=40

Step-by-step explanation:

5/8 as well as 10/16 = .625

a=8 blocks for 5 min given in question

b:24*0.625=15

c:25*40=.625 or 8 * 5=40 , since 5×5=25 taking 1st column times 5

3 0
3 years ago
You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 nigh
kirill [66]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean GPA of night students and μ2 be the mean GPA of day students.

The random variable is μ1 - μ2 = difference in the mean GPA of night students and the mean GPA of day 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

This is a two tailed test.

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

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 2.35

μ2 = 2.58

s1 = 0.46

s2 = 0.47

n1 = 30

n2 = 25

t = (2.35 - 2.58)/√(0.46²/30 + 0.47²/25)

t = - 1.8246

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 = [0.46²/30 + 0.47²/25]²/[(1/30 - 1)(0.46²/30)² + (1/25 - 1)(0.47²/25)²] = 0.00025247091/0.00000496862

df = 51

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

p value = 0.0746

Since alpha, 0.05 < than the p value, 0.0746, then we would fail to reject the null hypothesis.

3 0
3 years ago
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