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3 years ago

Which equation has the solution n = 11?

Mathematics

2 answers:

yawa3891 [41]3 years ago

6 0

Answer:

3n + 12 = 45

Step-by-step explanation:

*Process of elimination*

___________________

3n + 12 = 45

(given)

_________

3n + 12 = 45

- 12 - 12

(subtraction property of equality)

__________________________

3n = 33

÷ 3 ÷ 3

(division property of equality)

_______________________

n = 11

_____

7n - 8 = 62

(given)

________

7n - 8 = 62

+ 8 + 8

(addition property of equality)

__________________________

7n = 70

÷ 7 ÷ 7

(division property of equality)

_______________________

n = 10 ≠ 11

________

8n + 6 = 102

(given)

________

8n + 6 = 102

- 6 - 6

(subtraction property of equality)

__________________________

8n = 96

÷8 ÷8

(division property of equality)

_______________________

n = 12 ≠ 11

________

5n - 20 = 25

(given)

________

5n - 20 = 25

+ 20 + 20

(addition property of equality)

__________________________

5n = 45

÷5 ÷5

(division property of equality)

_______________________

n = 9 ≠ 11

kap26 [50]3 years ago

4 0

Answer:

3n + 12 = 45

Step-by-step explanation:

3n + 12 = 45

3n = 45 - 12

3n = 33

n = 33

n = 11

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Step-by-step explanation:

Let the no of 2kg disks be a.

Let no of 5 kg disks be b.

2a+5b=29

a+b=10

Using substitution, we have

2(10-b) + 5b=29

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20+3b=29

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Answer:

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$n=10$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =0.0217$ represent the sample variance

$\sigma^2_0 =0.12^2= 0.0144$ represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 0.0144$

Alternative hypothesis: $\sigma^2 \neq 0.0144$

The statistic can be calculated like this;

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

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