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Brilliant_brown [7]
3 years ago
15

Evaluate. 17 + 330 • 20 ÷

lign="absmiddle" class="latex-formula"> ÷ 2 – 14
Mathematics
1 answer:
Svetlanka [38]3 years ago
6 0

Answer:

36

Step-by-step explanation:

Simplify the following:

17 + 330×20/10^2×1/2 - 14

20/10^2×1/2 = 20/(10^2×2):

17 + 330×20/(10^2×2) - 14

330×20/(10^2×2) = (330×20)/(10^2×2):

17 + (330×20)/(10^2×2) - 14

2 | 1 | 6 | 5

| 3 | 3 | 0

- | 2 | |  

| 1 | 3 |  

- | 1 | 2 |  

| | 1 | 0

| - | 1 | 0

| | | 0:

17 + (165×20)/10^2 - 14

| 1 | 0

× | 1 | 0

| 0 | 0

1 | 0 | 0

1 | 0 | 0:

17 + (165×20)/100 - 14

20/100 = 20/(20×5) = 1/5:

17 + 165/5 - 14

The gcd of 165 and 5 is 5, so 165/5 = (5×33)/(5×1) = 5/5×33 = 33:

17 + 33 - 14

| 1 |  

| 3 | 3

+ | 1 | 7

| 5 | 0:

50 - 14

| 4 | 10

| 5 | 0

- | 1 | 4

| 3 | 6:

Answer:  36

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Answer:

t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802  

The degrees of freedom are given by:

df=n_{1}+n_{2}-2=4+4-2=6

The p value for this case would be given by:

p_v =2*P(t_{(6)}

Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

Step-by-step explanation:

Information given

Battery 1 106 111 109 105

Battery 2 125 103 121 118

We can calculate the mean and the deviation with the following formulas"

\bar X =\frac{\sum_{i=1}^n X_i}{n}

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

\bar X_{1}=107.75 represent the mean for the Battery 1

\bar X_{2}=116.75 represent the mean for the Bettery 2

s_{1}=2.75 represent the sample standard deviation for the Battery 1

s_{2}=9.604 represent the sample standard deviation for the battery 2

n_{1}=4 sample size selected for the Battery 1

n_{2}=4 sample size selected for the Battery 2

\alpha=0.1 represent the significance level

t would represent the statistic  

p_v represent the p value

System of hypothesis

We want to check if the difference in longevity between the two batteries, the system of hypothesis would be:

Null hypothesis:\mu_{1} = \mu_{2}

Alternative hypothesis:\mu_{1} \neq \mu_{2}

The statistic is given by:

t=\frac{\bar X_{s1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

The statistic is given by:

t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802  

The degrees of freedom are given by:

df=n_{1}+n_{2}-2=4+4-2=6

The p value for this case would be given by:

p_v =2*P(t_{(6)}

Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

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