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4 years ago

Any One Know This One

Mathematics

2 answers:

oksian1 [2.3K]4 years ago

4 0

Answer:

no solutions

Step-by-step explanation:

4 + 5m = 4m + 1 + m combine like terms

4 + 5m = 5m + 1 subtract 5m from each side

4 = 1

4 can never equal 1, so no value of m would allow the equation to be correct

artcher [175]4 years ago

3 0

<h3>Answer: C) No solutions</h3>

The left side 4+5m is the same as 5m+4

Simplifying the right hand side has 4m+1+m turn into 5m+1

The original equation is equivalent to 5m+4 = 5m+1

If we subtract 5m from both sides, the 5m terms cancel leaving us with 4 = 1, but that's not a true statement.

Regardless of whatever value you pick for m, the original equation will not be true.

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lara31 [8.8K]

Answer:

Step-by-step explanation:

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8 0

3 years ago

Is it A B C D i need an anwer asap

kompoz [17]

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4 years ago

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The

adelina 88 [10]

Answer:

The 90% confidence level is $19.15< L < 20.85$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 64$

The mean age is $\= x = 20 \ years$

The standard deviation is $\sigma = 4 \ years$

Generally the degree of freedom for this data set is mathematically represented as

$df = n - 1$

substituting values

$df = 64 - 1$

$df = 63$

Given that the level of confidence is 90% the significance level is mathematically evaluated as

$\alpha = 100 - 90$

$\alpha =$ 10 %

$\alpha = 0.10$

Now $\frac{\alpha }{2} = \frac{0.10}{2} = 0.05$

Since we are considering a on tail experiment

The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

$t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669$

The margin for error is mathematically represented as

$MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values

$MOE = 1.699 * \frac{4 }{\sqrt{64} }$

$MOE = 0.85$

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

$\= x - MOE < L < \= x + E$

substituting values

$20 - 0. 85 < L < 20 + 0.85$

$19.15< L < 20.85$

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3 years ago

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melisa1 [442]

Your answer is B because the amount needs to be greater or equal to $1,500

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3 years ago

Solve the linear equation. 142 - (41 + 6) = 3 (x – 4) – 18

Dvinal [7]

Answer:

X=41.667

Step-by-step explanation

142-(41+6)=3(X-4)-18

142-47=3x-12-18

95=3x-30

3x=125

x= $\frac{125}{3}$

x=41.667

3 0

4 years ago