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4 years ago

Is this an expression? n+6=20 Yes or No

Mathematics

2 answers:

Ainat [17]4 years ago

4 0

Yes it is an expression

Ghella [55]4 years ago

3 0

An expression consists of one or more numbers and variables along with one or more operations. However, it does not have an equals sign.

n + 6 = 20 would be an equation, not an expression.

An equation is a mathematical sentence that has an equal sign.

Therefore, n + 6 = 20 would not be an expression.

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Which restriction is appropriate for the variable 4a + 12 / a + 3

bonufazy [111]

I'll presume the slash is the fraction bar and the questioner is asking about

$\dfrac{4a+12}{a+3}$

That always equals 4 except when $a=-3$

So the restriction is

Answer: a ≠ -3

5 0

3 years ago

Read 2 more answers

What is 43 percent of 84

Misha Larkins [42]

Answer:

36.12

Step-by-step explanation:

43% of 84 can be rewritten to 43% * 84 or 43/100 * 84.

43/100 * 84/1 = 3612/100

3612/100 = 36.12

3 0

3 years ago

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X is all of the following except ____.

den301095 [7]

I’m pretty sure it’s D

3 0

3 years ago

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Find the equation of the line passing through the points (3,-2) and (4, 6). y = 8x + [?]

Sauron [17]

Step-by-step explanation:

1.) Plug in one of the points.

y = 8x + b

6 = 8(4) + b

2.) Solve for b (or [?])

$6 = 32 + b \\ - 32 \: \: \: \: \: \: \: \: \: - 32$

32 cancels out on the right side.

Subtract 32 from 6, which is -26

3.) -26 = b

4.) Check with other point:

-2 = 8(3) + b

-2 = 24 + b

Cancel out 24 on the right side: 24-24 = 0

-2 - 24 = -26

-26 = b

4 0

3 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago