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3 years ago

A number N is multiplied by 3. The result is the same as when N is divided by 3. What is the value of N?

Mathematics

2 answers:

nignag [31]3 years ago

8 0

The answer MIGHT be 0, but 0/3 is also undefined.

Otrada [13]3 years ago

3 0

Answer:

$N=0$

Step-by-step explanation:

We have been given that a number N is multiplied by 3. So our number would be $3\cdot N$ .

We are also told that N is divided by 3. So our number would be $\frac{N}{3}$ .

As we are told that the both results are same, so we can equate both expressions as:

$3\cdot N=\frac{N}{3}$

$3*3\cdot N=\frac{N}{3}*3$

$9N=N$

$9N-N=N-N$

$8N=0$

$\frac{8N}{N}=\frac{0}{N}$

$N=0$

Therefore, the value of N is 0.

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The sum of 3 consecutive integers is 48. What is the smallest of the integers?

Mademuasel [1]

Answer:

Step-by-step explanation:

x+(x+1)+(x+2)= 48

=)3x+3=48

=)3x=45

X=15

8 0

3 years ago

By law, an industrial plant can discharge not more than 500 gallons of waste water per hour, on the average, into a neighboring

Alisiya [41]

Answer:

We conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.

Step-by-step explanation:

We are given that an industrial plant can discharge not more than 500 gallons of wastewater per hour, on average, into a neighboring lake.

Four one-hour periods are selected randomly over a period of one week. The following are observed:

1384, 683, 1534, 405

Let $\mu$ = population average gallons of wastewater discharged per hour

So, Null Hypothesis, $H_0$ : $\mu \leq$ 500 gallons {means that not more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}

Alternate Hypothesis, $H_A$ : $\mu$ > 500 gallons {means that more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

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

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 1001.5 gallons

s = sample standard deviation = $\sqrt{\frac{\sum (X - \bar X)^{2} }{n-1} }$ = 543.79

n = sample of periods = 4

So, the test statistics = $\frac{1001.5-500}{\frac{543.79}{\sqrt{4} } }$ ~ $t_3$

= 1.844

The value of t-test statistics is 1.844.

Since in the question we are not given with the level of significance so we assume it to be 5%. Now, at 0.05 level of significance, the t table gives a critical value of 2.353 at 3 degrees of freedom for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 1.844 < 2.353, so we have insufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.

4 0

3 years ago

Write two equivalent expressions to represent the perimeter of this rectangle.

Harlamova29_29 [7]

P= (2)(x-2)+(2)(5)
P= (x-2)+(x-2)+5+5

8 0

1 year ago

Read 2 more answers

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