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inna [77]
2 years ago
6

tle="\lim_{x \to 0 (\frac{x(x-2)}{2-2e^2x} )" alt="\lim_{x \to 0 (\frac{x(x-2)}{2-2e^2x} )" align="absmiddle" class="latex-formula">
Help evaluting this limit
Mathematics
1 answer:
krek1111 [17]2 years ago
7 0

Answer: 0

Step-by-step explanation:

Substituting in x=0, we get

\frac{0(0-2)}{2-2e^{2}(0)}=0

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Welppp fastttt!!!!!!!!!!!!!!!!!!
Romashka [77]

Answer: B

Step-by-step explanation:

Grade must be greater than or equal to 88

7 0
3 years ago
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Hey I’m struggling with number 13 so please show your work. Thx
Dafna11 [192]

Answer:

\frac{5}{12}

Step-by-step explanation:

Because \frac{3}{5} is being multiplied by S, we can divide \frac{3}{5} from both sides of the equation. This will give us:

\frac{1}{4}÷\frac{3}{5}

But, that looks a bit hectic. Instead of dividing, you can multiply by the reciprocal (which is essentially how to divide fractions). So, instead of \frac{1}{4}÷\frac{3}{5}, you get:

\frac{1}{4}×\frac{5}{3}

When multiplying fractions, remember you can just multiply straight across-- numerator x numerator, then denominator x denominator.

By doing that, you get the fraction \frac{5}{12}

\frac{5}{12} cannot be simplified any more, so S=\frac{5}{12} is your answer :)

I hope this helps

6 0
3 years ago
How do you solve (7-2)/5 step by step
gtnhenbr [62]
7 - 2 = 5

5 / 5 = 1

The answer would be 1.
3 0
3 years ago
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What is the solution to the equation?
Otrada [13]

Answer:

The format of this equation is wrong

3 0
3 years ago
An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberr
seraphim [82]

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 compression strength from the extra carbonation of strawberry drink and μ2 be the mean compression strength from the extra carbonation of cola.

The random variable is μ1 - μ2 = difference in the mean compression strength from the extra carbonation of strawberry drink and the mean compression strength from the extra carbonation of cola.

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 left 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,

x1 = 530

x2 = 553

s1 = 23

s2 = 16

n1 = 10

n2 = 10

t = (530 - 533)/√(23²/10 + 16²/10)

t = - 2.6

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 = [23²/10 + 16²/10]²/[(1/10 - 1)(23²/10)² + (1/10 - 1)(16²/10)²] = 6162.25/383.752

df = 16

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

p value = 0.0097

Since alpha, 0.05 > than the p value, 0.0097, then we would reject the null hypothesis. Therefore, at 5% significance level, the data suggests that the extra carbonation of cola results in a higher average compression strength.

4 0
3 years ago
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