All categories

2 years ago

17n + 4n- 4= 122 how do i solve for n?

Mathematics

1 answer:

Leokris [45]2 years ago

8 0

Answer:

n = 6

Step-by-step explanation:

17n + 4n - 4 =122

combine like terms 17n & 4n

21n - 4 = 122

21n - 4 + 4 = 122 + 4

21n = 126

21n / 21 = 126 / 21

n = 6

You might be interested in

What is point of view

melamori03 [73]

Answer:

"a particular attitude or way of considering a matter. "

hope this helps :3

4 0

2 years ago

726,000 scientific notation<br><br>

il63 [147K]

Answer:

7.26000 × $10^{5}$

I hope this helps!

3 0

2 years ago

Read 2 more answers

I need help with this please

Volgvan

Answer: No she is not.

Step-by-step explanation: The correct answer is 7.25 F, because she just put them both as positives. When you have a negative it takes away from the total, so how I think of it is you subtract from the positive number, so you could just do 11.75-4.5. Hope this helped!

7 0

2 years ago

FREE BARINLIEST

DerKrebs [107]

The answer would have to be 30%

3 0

2 years ago

Read 2 more answers

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batterie

Fittoniya [83]

Answer:

The confidence interval is $25.16 < \mu < 26.85$

Step-by-step explanation:

From the question we are given a data set

25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is

$\= x = \frac{\sum x_i}{n}$

where is the sample size with values n = 10

$\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}$

$\= x = 26$

The standard deviation is evaluated as

$\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }$

substituting values

$= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }$

$\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }$

$\sigma = 1.625$

The confidence level is given as 90% hence the level of significance is calculated as

$\alpha = 100 -90$

$\alpha =10$ %

$\alpha = 0.10$

Now the critical values of $\frac{\alpha }{2}$ is obtained from the normal distribution table as

$Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining the critical values of $\frac{\alpha }{2}$ instead of $\alpha$ is because we are considering two tails of the area under the normal curve